Math 1326 Notes

1.2.1 Definition

1.2.2 How to determine

1.3.1 Definition

1.3.2 Example

1.4.1 Arithmetic

1.4.2 Composition

• Definition
  
  • \((f\circ g) (x) = f(g(x))\)
  • \((g\circ f) (x) = g(f(x))\)
  • In general, \((f\circ g) (x) \neq (g\circ f) (x)\)
• How to find \(f\circ g\) for given \(f\) and \(g\)
  
  • Steps
    1. In \(f(x)\), replace variable \(x\) with variable \(g\), and
    2. Then replace variable \(g\) with the expression of \(g(x)\)
  • Examples

    • \(f(x) = 3, g(x) = 4\)

      \((f\circ g) (x) = f(g(x)) = f(4) = 3\)

    • \(f(x) = 3, g(x) = ln(x)\)

      \((f\circ g) (x) = f(g(x)) = f(ln(x)) = 3\)

    • \(f(x) = x^2 + 1, g(x) = \sqrt{x}\)

      \((f\circ g) (x) = f(g) = g^2 + 1 = [g(x)]^2 + 1 = (\sqrt{x})^2 + 1 = x + 1, x \geq 0\)

    • \(f(x) = ln(\frac{x^2}{1-x}), g(x) = 4^{x+1}\)

      \((f\circ g) (x) = f(g) = ln(\frac{g^2}{1-g}) = ln(\frac{[g(x)]^2}{1-g(x)}) = ln(\frac{[4^{x+1}]^2}{1-4^{x+1}})\)

      \((g\circ f) (x) = g(f) = 4^{f+1} = 4^{f(x) + 1} = 4^{ln(\frac{x^2}{1-x}) + 1}\)

• How to identify \(f\) and \(g\) for a given composition
  
  • Steps
    • Let \(f\) be the outermost function, so the composition can be written as \(f(g)\), a function of \(g\)
    • Let \(g\) be all terms immediately inside \(f\), so \(g\) can be written as \(g(x)\), a function of \(x\)
  • Examples

    • \(h(x) = ln(3x+1)\)

      \(ln(\text{something})\) is the outermost function, so let \(f(g) = ln(g)\)

      The whole term immediately inside \(ln()\) function is \(g(x) = 3x + 1\)

      Thus \(f(x) = ln (x)\) and \(g(x)= 3x + 1\)

    • \(h(x) = (7x^2 + 3x - 2)^{100}\)

      \((\text{something})^{100}\) is the outermost function, so let \(f(g) = g^{100}\)

      The whole term immediately inside \((\cdot)^{100}\) is \(g(x) = 7x^2 + 3x - 2\)

      Thus \(f(x) = x^{100}\) and \(g(x) = 7x^2 + 3x - 2\)

    • \(h(x) = 3(x-2)^2 + 5\)

      \(3 \cdot \text{something} + 5\) is the outermost function, so let \(f(g) = 3g + 5\) and \(g(x) = (x-2)^2\)

      Or, regard \(3(\text{something})^2 + 5\) as the outermost function, so \(f(g) = 3g^2 + 5\) and \(g(x) = x-2\)

      Or, regard \(\text{something} + 5\) as the outermost function, so \(f(g) = g + 5\) and \(g(x) = 3(x-2)^2\)

1.5.1 Definition

1.5.2 Property

1.6.1 Defintion

1.6.2 Properties

1.6.3 Application of Logarithmic Properties

2.3.1 Arithmetic Rules

2.3.2 Important Results

2.4.1 Technique 1 : Substitution

• Polynomial
  

  \(\lim\limits_{x\to 2} x^3 - 3x^2 + 5x - 2 = 2^3 - 3\cdot 2^2 + 5\cdot 2 - 2 = 4\)

• Fraction with Denominator Non-Zero
  

  \(\lim\limits_{x\to 2} \frac{2x}{3x^3+1} = \frac{2\cdot 2}{3\cdot 2^3 + 1} = \frac{4}{25}\)

• Square Root Function
  

  \(\lim\limits_{x\to 2} \sqrt{x^2 + 5} = \sqrt{2^2 + 5} = 3\)

• Exponential Function
  

  \(\lim\limits_{x\to 2} 3^{x+1} = 3^{2 + 1} = 3^3 =27\)

• Logarithmic Function
  

  \(\lim\limits_{x\to 2} log_2 x^5 = log_2 2^5 = 5\)

2.4.2 Technique 2 : Factorization and Reduction

• Fraction of Polynomials with Denominator Zero
  

  \(\lim\limits_{x\to 3} \frac{2x^2 - 5x - 3}{x-3}\).

  If we directly plug \(3\) into the function, we get denominator \(0\). Hence we have to factorize the numerator and then cancel some common factor.

  \(\lim\limits_{x\to 3} \frac{2x^2 - 5x - 3}{x-3}\) \(= \lim\limits_{x\to 3} \frac{(2x + 1)(x-3)}{x-3}\) \(= \lim\limits_{x\to 3} 2x + 1 = 2\cdot 3 + 1 = 7\).

2.4.3 Technique 3 : Multiplication by Conjugate

• Fraction Containing Square Root with Denominator Zero
  

  \(\lim\limits_{x\to 9} \frac{\sqrt{x} - 3}{x-9}\).

  If we directly plug \(3\) into the function, we get denominator \(0\). Also, neither the numerator nor the denominator can be factorzied. However, We note the numerator is a square root function. In this case, we multiply both numerator and denominator by the conjugate of the numerator (because the numerator contains square root symbol. Similary, if the denominator contains square root symbol, then we multiply by the conjugate of the denominator). The conjuate of \(\sqrt{x} - 3\) is \(\sqrt{x} + 3\), so we have

  \(\lim\limits_{x\to 9} \frac{\sqrt{x} - 3}{x-9}\)

  \(= \lim\limits_{x\to 9} \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x-9)(\sqrt{x} + 3)}\) \(= \lim\limits_{x\to 9} \frac{(\sqrt{x})^2 - 3^2}{(x-9)(\sqrt{x} + 3)}\) \(= \lim\limits_{x\to 9} \frac{x - 9}{(x-9)(\sqrt{x} + 3)}\)

  \(= \lim\limits_{x\to 9} \frac{1}{\sqrt{x} + 3}\) \(= \frac{1}{6}\).

2.4.4 Technique 4 : Limit at Infinity of a Fraction

• Case 1 : Limit at Infinity of a Fraction Without Square Root
  

  • Description

    Suppose \(f(x) = \frac{g(x)}{h(x)}\) where

    • \(g(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) is a polynomial of order \(p\)
    • \(h(x) = b_0 + b_1 x + b_2 x^2 + \dots + b_q x^q\) is a polynomial of order \(q\)

    To find \(\lim\limits_{x\to\infty} f(x)\) or \(\lim\limits_{x\to -\infty} f(x)\), we need

    • Divide both numerator and denominator by \(x^{max\{p, q\}}\) and simplify the quotients, and
    • Evaluate the limit using the simplified form.

  • Example

    \(\lim\limits_{x\to \infty} \frac{2x^2 -3x + 1}{x^2 -5}\)

    \(= \lim\limits_{x\to \infty} \frac{2x^2 -3x + 1}{x^2 -5}\) \(= \lim\limits_{x\to \infty} \frac{\frac{2x^2 -3x + 1}{x^2}}{\frac{x^2 -5}{x^2}}\) \(= \lim\limits_{x\to \infty} \frac{2 - \frac{3}{x} + \frac{1}{x^2}} {1 - \frac{5}{x^2}}\)

    \(= \frac{ \lim\limits_{x\to \infty} 2 - \lim\limits_{x\to \infty} \frac{3}{x} + \lim\limits_{x\to \infty} \frac{1}{x^2}} {\lim\limits_{x\to \infty} 1 - \lim\limits_{x\to \infty} \frac{5}{x^2}}\)

    \(= \frac{2-0+0}{1-0}\) \(= 2\)

  • Remarks : Shortcuts
    • If \(p=q\), then \(\lim\limits_{x\to\infty} f(x) = \lim\limits_{x\to-\infty} f(x) = \frac{a_p}{b_q}\).

    • If \(p < q\), then \(\lim\limits_{x\to\infty} f(x) = \lim\limits_{x\to-\infty} f(x) = 0\).

      \(\lim\limits_{x\to\infty} \frac{x^2+1}{x^3-2} = 0\)

    • If \(p > q\), then

      • \(\lim\limits_{x\to\infty} f(x) = \infty\).

        \(\lim\limits_{x\to\infty} \frac{x^3-2}{x^2+1} = \infty\)

      • \(\lim\limits_{x\to -\infty} f(x) =\)

        • \(\infty\) if \(p-q\) is even
        • \(-\infty\) if \(p-q\) is odd.

        \(\lim\limits_{x\to -\infty} \frac{x^3-2}{x^2+1} = -\infty\)

        \(\lim\limits_{x\to -\infty} \frac{x^4-2}{x^2+1} = \infty\)

• Case 2 : Limit at Infinity of a Fraction With Square Root
  

  • Description

    Suppose

    • \(g(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) is a polynomial of order \(p\)
    • \(h(x) = b_0 + b_1 x + b_2 x^2 + \dots + b_q x^q\) is a polynomial of order \(q\)

    Then

    • If \(f(x) = \frac {\sqrt{g(x)}} {h(x)}\), then divide both numerator and denominator by \(max\{p/2, q\}\).
    • If \(f(x) = \frac {g(x)} {\sqrt{h(x)}}\), then divide both numerator and denominator by \(max\{p, q/2\}\).
    • If \(f(x) = \frac {\sqrt{g(x)}} {\sqrt{h(x)}}\), then divide both numerator and denominator by \(max\{p/2, q/2\}\).

  • Example

    \(\lim\limits_{x\to \infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

    Now we can see

    • \(f(x) = \frac {\sqrt{g(x)}} {h(x)}\)
    • \(p = 2\)
    • \(q = 1\)
    • \(max\{p/2, q\} = max\{ 2/2, 1 \} = 1\)

    so we divide both top and bottom by \(x^1 = x\) :

    \(\lim\limits_{x\to \infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

    \(= \lim\limits_{x\to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{x -5}{x}}\) \(= \lim\limits_{x\to \infty} \frac {\frac {\sqrt{2x^2 + 1}} {\sqrt{x^2}} } {\frac {x-5} {x} }\) \(= \lim\limits_{x\to \infty} \frac {\sqrt{ \frac {2x^2 + 1} {x^2} } } {1 - \frac {5} {x} }\)

    \(= \lim\limits_{x\to \infty} \frac {\sqrt{ 2 + \frac{1}{x^2} } } {1 - \frac {5} {x} }\) \(= \frac {\lim\limits_{x\to \infty} \sqrt{ 2 + \frac{1}{x^2} } } {\lim\limits_{x\to \infty} 1 - \frac {5} {x} }\) \(= \frac{\sqrt{2+0}}{1-0}\)

    \(= \sqrt{2}\)

    Here we use \(x = \sqrt{x^2}\), which is only true for \(x\geq 0\). For \(x < 0\), \(x = -\sqrt{x^2}\) holds. Since here we are evaluating the limit at \(+\infty\), we consider only positive \(x\). If we evaluate the limit at \(-\infty\), then it should be

    \(\lim\limits_{x\to -\infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

    \(= \lim\limits_{x\to -\infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{x -5}{x}}\) \(= \lim\limits_{x\to -\infty} \frac {\frac {\sqrt{2x^2 + 1}} {-\sqrt{x^2}} } {\frac {x-5} {x} }\) \(= \lim\limits_{x\to -\infty} - \frac {\sqrt{ \frac {2x^2 + 1} {x^2} } } {1 - \frac {5} {x} }\)

    \(= \lim\limits_{x\to -\infty} -\frac {\sqrt{ 2 + \frac{1}{x^2} } } {1 - \frac {5} {x} }\) \(= -\frac {\lim\limits_{x\to -\infty} \sqrt{ 2 + \frac{1}{x^2} } } {\lim\limits_{x\to -\infty} 1 - \frac {5} {x} }\) \(= -\frac{\sqrt{2+0}}{1-0}\)

    \(= -\sqrt{2}\)

3.1.1 Formal Definition

3.1.2 Find Derivatives by Definition

• Steps
  
  1. Find \(f(x + h)\)
  2. Find and simplify \(f(x + h) - f(x)\)
  3. divide by \(h\) to get \(\frac{f(x + h) - f(x)} {h}\)
  4. Let \(h\to 0\) : \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\), if the limit exists.
• Examples
  

  • Example 1 : Constant Function

    \(f(x) = 10\). Find \(f'(x)\).

    \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {10 - 10} {h}\)

    \(= \lim\limits_{h \to 0} 0\)

    \(= 0\)

  • Example 2 : Polynomial Function

    \(f(x) = 3x^2 + 5x - 1\). Find \(f'(3)\).

    \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {[3(x+h)^2 + 5(x+h) - 1] - (3x^2 + 5x - 1)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {[3(x^2 + h^2 + 2xh) + 5(x+h) -1] - (3x^2 + 5x - 1)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {3x^2 + 3h^2 + 6xh + 5x + 5h -1 - 3x^2 - 5x + 1} {h}\)

    \(= \lim\limits_{h \to 0} \frac {3h^2 + 6xh + 5h} {h}\)

    \(= \lim\limits_{h \to 0} 3h + 6x + 5\)

    \(= 6x + 5\)

    Note:

    To find \(f'(a)\), first find \(f'(x)\), and then substitute \(x\) by \(a\) in \(f'(x)\). The order can't be changed.

    • RIGHT

      \(f'(x) = 6x + 5\), so subsitute \(x\) by \(3\) and get \(f'(3) = 6\cdot 3 + 5 = 23\)

    • WRONG

      \(f(3) = 3\cdot 3^2 + 5\cdot 3 - 1 = 41\), which is a constant, so \(f'(3) = 0\).

  • Example 3 : Sqrare Root Function : Using Conjugate

    \(f(x) = \sqrt{2x - 1} - 7\). Find \(f'(x)\).

    \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {(\sqrt{2(x + h) - 1} - 7) - (\sqrt{2x - 1} - 7)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {\sqrt{2(x + h) - 1} - \sqrt{2x - 1}} {h}\)

    \(= \lim\limits_{h \to 0} \frac {\sqrt{2(x + h) - 1} - \sqrt{2x - 1}} {h} \cdot \frac {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

    \(= \lim\limits_{h \to 0} \frac {[2(x + h) - 1] - (2x - 1)} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

    \(= \lim\limits_{h \to 0} \frac {2x + 2h - 1 - 2x + 1} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

    \(= \lim\limits_{h \to 0} \frac {2h} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

    \(= \lim\limits_{h \to 0} \frac {2} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

    \(= \frac{2}{\sqrt{2x - 1} + \sqrt{2x - 1}}\)

    \(= \frac{2}{2\sqrt{2x - 1}}\)

    \(= \frac{1}{\sqrt{2x - 1}}\)

    Note:

    • Conjugate of \(\sqrt{2(x + h) - 1} - \sqrt{2x - 1}\) is \(\sqrt{2(x + h) - 1} + \sqrt{2x - 1}\).
    • \(\sqrt{2(x + h) - 1} - \sqrt{2x - 1} \neq \sqrt{2h}\)

  • Example 4 : Fraction : Using Common Denominator

    \(f(x) = \frac{2}{3x+1}\). Find \(f'(x)\).

    \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

    \(= \lim\limits_{h \to 0} \frac {\frac{2}{3(x+h)+1} - \frac{2}{3x+1}} {h}\) \(= \lim\limits_{h \to 0} \frac {\frac{2\cdot (3x+1)}{[3(x+h)+1] \cdot (3x+1)} - \frac{2\cdot[3(x+h)+1]}{[3(x+h)+1] \cdot (3x+1)}} {h}\) \(= \lim\limits_{h \to 0} \frac {\frac {2\cdot (3x+1) - 2\cdot[3(x+h)+1]} {[3(x+h)+1] \cdot (3x+1)}} {h}\)

    \(= \lim\limits_{h \to 0} \frac{2\cdot (3x+1) - 2\cdot[3(x+h)+1]} {h\cdot [3(x+h)+1] \cdot (3x+1)}\) \(= \lim\limits_{h \to 0} \frac{6x + 2 - 6x - 6h - 2} {h\cdot [3(x+h)+1] \cdot (3x+1)}\) \(= \lim\limits_{h \to 0} \frac{- 6h} {h\cdot [3(x+h)+1] \cdot (3x+1)}\)

    \(= \lim\limits_{h \to 0} \frac{- 6} {[3(x+h)+1] \cdot (3x+1)}\) \(= \frac{- 6} {(3x+1) \cdot (3x+1)}\) \(= \frac{- 6} {(3x+1)^2}\)

3.2.1 Arithmetic Rules

• Addition & Subtraction
  

  If \(f(x) = g(x) \pm h(x)\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[ f'(x) = g'(x) \pm h'(x) \]

• Scalar Multiplication
  

  If \(f(x) = k g(x)\) where \(k\) is a real number and \(g'(x)\) exists, then \[ f'(x) = k g'(x) \]

• Product Rule
  

  If \(f(x) = g(x) \cdot h(x)\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[ f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)\]

  • Example

    \(f(x) = g(x) \cdot h(x)\) where \(g(x) = 2x^3 + 1\) and \(h(x) = 7x^2 - 2\)

    \(g'(x) = 2\cdot 3\cdot x^2 = 6x^2\)

    \(h'(x) = 7\cdot 2\cdot x = 14x\)

    \(f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)\)

    \(= 6x^2 (7x^2 - 2) + (2x^3 + 1) \cdot 14x\)

    \(= 42 x^4 - 12 x^2 + 28 x^4 + 14x\)

    \(= 70 x^4 - 12 x^2 + 14x\)

• Quotient Rule
  

  If \(f(x) = \frac {g(x)} {h(x)}\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[f'(x) = \frac {g'(x) \cdot h(x) - g(x) \cdot h'(x)} {[h(x)]^2} \]

  • Example

    \(f(x) = \frac{g(x)} {h(x)}\) where \(g(x) = 2x^3 + 1\) and \(h(x) = 7x^2 - 2\)

    \(g'(x) = 2\cdot 3\cdot x^2 = 6x^2\)

    \(h'(x) = 7\cdot 2\cdot x = 14x\)

    \(f'(x) = \frac {g'(x) \cdot h(x) - g(x) \cdot h'(x)} {[h(x)]^2}\)

    \(= \frac {6x^2 \cdot (7x^2 - 2) - (2x^3+1) \cdot 14x} {(7x^2-2)^2}\)

    \(= \frac {6x^2 (7x^2 - 2) - (2x^3 + 1) \cdot 14x} {(7x^2-2)^2}\)

    \(= \frac {42 x^4 - 12 x^2 - 28 x^4 - 14x} {(7x^2-2)^2}\)

    \(= \frac {14 x^4 - 12 x^2 - 14x} {(7x^2-2)^2}\)

3.2.2 Chain Rule for Composite Function

• Statement
  

  For a composite function \(f(g(x))\), if we regard the function \(g(x)\) as a single variable \(g\) by letting \(g = g(x)\), then \[ \frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} \]

• Remarks
  
  • \(g\) and \(g(x)\)
    • when writting write \(g\), we mean a variable \(g\), just like \(x, u\) and \(v\)
    • when writing \(g(x)\), we mean a function of \(x\)
  • \(\frac{df(g)}{dg}\) means
    • the variable of function \(f(g)\) is \(g\)
    • The derivative of \(f(g)\) is with respect to the variable \(g\)
  • In short notation, we may memorize the chain rule by (plausibly, you get the LHS by cancelling \(dg\) on the RHS) \[ \frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]
• Example
  

  Let \(f(x) = (2x^2 + 1)^3\). Find \(\frac{df(x)}{dx}\).

  Clearly \(f(x)\) is a composite function, where

  • The inner function is \(g(x) = 2x^2 + 1\).
  • The outer function is \(f(g) = g^3\) where \(g=g(x)\).

  By Chain Rule \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx}\), to find \(\frac{df(g(x))}{dx}\), we need \(\frac{df(g)}{dg}\) and \(\frac{dg(x)}{dx}\).

  • By regarding \(g\) as a single variable, we find \(\frac{df(g)}{dg} = \frac{d(g^3)}{dg} = 3 g^2\) by power rule
  • By regarding \(x\) as the variable, we find \(\frac{dg(x)}{dx} = \frac{d(2x^2 + 1)}{dx} = 4x\) by power rule
  • Putting them into the formula of chain rule, we get \(\frac{df(g(x))}{dx} = 3 g^2 \cdot 4x = 12xg^2\)
  • Here \(g\) is a variable that is not given by the question, so we have to replace it. Since \(g = g(x)\), we just replace \(g\) with the expression of \(g(x)\), which is \(2x^2 + 1\) here. Thus the final answer is \(\frac{df(g(x))}{dx} = 12x(2x^2+1)^2\).

3.2.3 Rules for Special Functions

• Constant Rule
  

  If \(f(x) = k\) where \(k\) is a real number, then \(f'(x) = 0\).

• Power Rule
  

  If \(f(x) = x^n\) where n is a real number, then \(f'(x) = n\cdot x^{n-1}\)

  /Remark: / Whenever you meet something like \(\frac{1}{x}\) or \(\sqrt{x}\), rewrite them as \(x^{-1}\) or \(x^{\frac{1}{2}}\) respectively.

• Polynomial Rule
  

  By the power rule and addition rule, we get that for a polynomial \(f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) where \(a_0, a_1, a_2, \dots, a_p\) are constants, the derivative is

  \(f'(x) = (a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p)'\)

  \(= (a_0)' + (a_1 x)' + (a_2 x^2)' + \dots + (a_p x^p)'\)

  \(= 0 + a_1 + a_2 \cdot 2x + a_3 \cdot 3x^2 + \dots + a_p \cdot px^{p-1}\)

  /Example: / \((5x^3 + 7x^2 - 3x + 2)' = 5 \cdot 3 x^2 + 7 \cdot 2x - 3\).

• Exponential Rule
  

  For any constant \(a\) such that \(a>0\) and \(a\neq 1\), \[ \frac{d}{dx} a^x = lna \cdot a^x \] \[ \frac{d}{dx} e^x = e^x \]

  • Examples
    • \(\frac{d}{dx} 3^x = ln3 \cdot 3^x\)

    • \(\frac{d}{dx} 3^{2x^3} =\) ?

      Let \(f(x) = 3^{2x^3}\), then we can see it is a composite function where

      • the inner function \(g(x) = 2x^3\), with \(\frac{dg(x)}{dx} = 6x^2\)
      • the outer function \(f(g) = 3^g\), with \(\frac{df(g)}{dg} = ln3 \cdot 3^g\) where \(g = g(x)\)

      so we should apply the Chain Rule and get

      \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = ln3 \cdot 3^g \cdot 6x^2 = 6 \cdot ln3 \cdot x^2 \cdot 3^{2x^3}\)

    • \(\frac{d}{dx} e^{4x^3 + 2} =\) ?

      Let \(f(x) = e^{4x^3 + 2}\), then we can see it is a composite function where

      • the inner function \(g(x) = 4x^3 + 2\), with \(\frac{dg(x)}{dx} = 4 \cdot 3x^2 + 0 = 12 x^2\)
      • the outer function \(f(g) = e^g\), with \(\frac{df(g)}{dg} = e^g\) where \(g = g(x)\)

      so we should apply the Chain Rule and get

      \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = e^g \cdot 12x^2 = e^{4x^3 + 2} 12x^2\)

    • \(\frac{d}{dx}e^2=\) ?

      Caution : \(e^2\) is a constant, so \(\frac{de^2}{dx} = e^2\) is WRONG. The exponential rule does not apply because \(2\) is not the variable \(x\). Instead, we should use the constant rule to get \(\frac{de^2}{dx} = 0\).

• Logarithmic Rule
  

  For any constant \(a\) such that \(a>0\) and \(a\neq 1\), \[ \frac{d}{dx} log_ax = \frac{1}{x \cdot lna} \] \[ \frac{d}{dx} lnx = \frac{1}{x} \]

  • Examples
    • \(\frac{d}{dx} log_2 x = \frac{1}{x \cdot ln2}\)

    • \(\frac{d}{dx} log_2 x^3 =\) ?

      Let \(f(x) = log_2 x^3\), then we can see it is a composite function where

      • the inner function \(g(x) = x^3\), with \(\frac{dg(x)}{dx} = 3x^2\)
      • the outer function \(f(g) = log_2 g\), with \(\frac{f(g)}{dg} = \frac{1}{g \cdot ln2}\)

      so we should apply the Chain Rule and get

      \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = \frac{1}{g \cdot ln2} \cdot 3x^2 = \frac{3x^2}{x^3 \cdot ln2}\)

    • \(\frac{d}{dx} ln(3x^2 - 1)=\) ?

      Let \(f(x) = ln (3x^2 - 1)\), then we can see it is a composite function where

      • the inner function \(g(x) = 3x^2 - 1\), with \(\frac{dg(x)}{dx} = 3\cdot 2x - 0 = 6x\)
      • the outer function \(f(g) = ln(g)\), with \(\frac{f(g)}{dg} = \frac{1}{g}\)

      so we should apply the Chain Rule and get

      \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = \frac{1}{g} \cdot 6x = \frac{6x}{3x^2 - 1}\)

    • \(\frac{d}{dx}ln2=\) ?

      Caution : \(ln2\) is a constant, so \(\frac{dln2}{dx} = \frac{1}{2}\) is WRONG. The logarithmic rule does not apply because \(2\) is not the variable \(x\). Instead, we should use the constant rule to get \(\frac{dln2}{dx} = 0\).

  • Remark

    Comparison between Power rule and Exponential Rule

    • Power Rule : \((x^a)' = a\cdot x^{a-1}\) requires variable \(x\) be the base and constant \(a\) be the power
    • Exponential Rule : \(a^x = lna \cdot a^x\) requires constant \(a\) \((a>0\) and \(a\neq 1)\) be the base and variable \(x\) be the power

3.3.1 Examples

• Example 1 : Using Logarithmic Identities
  

  For example, if \(f(x) = ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\), then we may use the logarithmic property to rewrite it as

  \(f(x) = ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\)

  \(= ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\)

  \(= ln(e^{3x^2 + 1}(2x+5)^2 \cdot \sqrt{3x-1}) - ln((4x+1)^3)\)

  \(= ln(e^{3x^2 + 1}) + ln((2x+5)^2) + ln(\sqrt{3x-1}) - ln((4x+1)^3)\)

  \(= (3x^2 + 1) lne + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)\)

  \(= 3x^2 + 1 + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)\)

  Now it is much easier for us to find the derivative.

  Using relevant rules, we get

  \(\frac{d}{dx} f(x)\)

  \(= \frac{d}{dx} [3x^2 + 1 + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)]\)

  \(= 3 \frac{d}{dx} x^2 + \frac{d1}{dx} + 2 \frac{d}{dx} ln(2x+5) + \frac{1}{2} \frac{d}{dx} ln(3x-1) - 3 \frac{d}{dx} ln(4x+1)\)

  \(= 3\cdot 2x + 0 + 2 \cdot \frac{2}{2x+5} + \frac{1}{2} \cdot \frac{3}{3x-1} - 3 \cdot \frac{4}{4x+1}\)

  \(= 6x + \frac{4}{2x+5} + \frac{3}{2(3x-1)} - \frac{12}{4x+1}\)

• Example 2 : Splitting Long Numerators
  

  For example, it is not a good idea to directly use quotient rule when we find the derivative of \(f(x) = \frac{x^3 + 2x^2 + 3x + 4}{5x^2}\). Instead, we may split the long numerator and then simplify each term:

  \(f(x) = \frac{x^3 + 2x^2 + 3x + 4}{5x^2}\)

  \(= \frac{x^3}{5x^2} + \frac{2x^2}{5x^2} + \frac{3x}{5x^2} + \frac{4}{5x^2}\)

  \(= \frac{x}{5} + \frac{2}{5} + \frac{3}{5}x^{-1} + \frac{4}{5}x^{-2}\)

  In this way, we immediately get the the derivative of \(f(x)\) as

  \(f'(x)\)

  \(= \left( \frac{x}{5} \right)' + \left( \frac{2}{5} \right)' + \left( \frac{3}{5} x^{-1} \right)' + \left( \frac{4}{5} x^{-2} \right)'\)

  \(= \frac{1}{5} + 0 + \frac{3}{5} \cdot (-1) x^{-2} + \frac{4}{5} \cdot (-2) x^{-3}\)

  \(= \frac{1}{5} - \frac{3}{5} x^{-2} - \frac{8}{5} x^{-3}\)

3.5.1 Definition

3.5.2 Critical Number and Critical Point

• Definitions
  

  The critical numbers for a function \(f\) are those numbers \(c\) in the domain of \(f\) for which

  • \(f'(c) = 0\) or
  • \(f'(c)\) does not exist

  A critical point is a point whose \(x\)-coordinate is the critical number \(c\) and whose \(y\)-coordinate is \(f(c)\).

• Example
  

  For \(f(x) = \frac{x^2 + 1}{2x - 1}\), by quotient rule we have

  \(f'(x)\)

  \(= \left( \frac{x^2 + 1}{2x - 1} \right)'\)

  \(= \frac{(x^2 + 1)'(2x - 1) - (x^2 + 1)(2x - 1)'}{(2x - 1)^2}\)

  \(= \frac{2x(2x - 1) - (x^2 + 1) \cdot 2}{(2x - 1)^2}\)

  \(= \frac{4x^2 - 2x - 2x^2 - 2}{(2x - 1)^2}\)

  \(= \frac{2x^2 - 2x - 2}{(2x - 1)^2}\)

  • \(f'(x)\) is undefined at \(x = \frac{1}{2}\), but \(\frac{1}{2}\) is NOT a critical number of \(f(x)\) because it is NOT in the domain

  • Letting \(f'(x) = 0\) means letting \(2x^2 - 2x - 2 = 0\). By formula for quadratic roots (the roots of \(ax^2 + bx + c = 0\) are \(x = \frac{-b \pm \sqrt{4ac}}{2a}\)), we know

    \(\frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-2)}}{2\cdot 2} = \frac{2 \pm \sqrt{20}}{4} = \frac{2 \pm 2 \sqrt{5}}{4} = \frac{1 \pm \sqrt{5}}{2}\) are 2 critical numbers.

  • Hence, only \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\) are critical numbers of \(f(x)\), and \(\left( \frac{1 + \sqrt{5}}{2}, f(\frac{1 + \sqrt{5}}{2}) \right)\) and \(\left( \frac{1 - \sqrt{5}}{2}, f(\frac{1 - \sqrt{5}}{2}) \right)\) are critical points. (Calculate \(f(\frac{1 - \sqrt{5}}{2})\) and \(f(\frac{1 + \sqrt{5}}{2})\) by yourself. The same for all following examples.)

3.5.3 For a Given Interval, How to Determine Whether Function \(f(x)\) is Increasing or Decreasing?

3.5.4 For a Given Function \(f(x)\), How to Find All Intervals on Which It Is Increasing or Decreasing?

• Procedure
  
  • Find all critical numbers of \(f(x)\)
  • Separate the whole domain of \(f(x)\) into several intervals by the critical numbers
  • Within each interval, choose a value of \(x\), say \(a\), and see whether \(f'(a) > 0, < 0\) or \(= 0\).
  • Decide where \(f(x)\) is increasing or decreasing on each interval
    • If \(f'(a) > 0\), then \(f(x)\) is increasing on that interval
    • If \(f'(a) < 0\), then \(f(x)\) is decreasing on that interval
    • If \(f'(a) = 0\), then \(f(x)\) is constant on that interval
• Example
  
  • We have already found that the \(f(x) = \frac{x^2 + 1}{2x - 1}\) is defined on \((-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\) and critical numbers of \(f(x)\) are \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\).
  • These critical numbers separate the domain of \(f(x)\) into the 1st row of intervals in the following table. Now we have to fill in the 2nd and the 3rd rows.
  • Within each interval, we choose a number \(a\) and evaluate \(f'(a)\) to see its sign

  • Based on the sign of \(f'(a)\), we determine whether the function is increasing or decreasing on that interval

    \(x\)	\((-\infty\), \(\frac{1 - \sqrt{5}}{2})\)	\((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\)	\((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\)	\((\frac{1 + \sqrt{5}}{2}, \infty)\)
    \(f'(x)\)	\(f'(-10) > 0\)	\(f'(0) < 0\)	\(f'(1) < 0\)	\(f'(10) > 0\)
    \(f(x)\)	increasing	desreasing	decreasing	increasing

3.6.1 Definition

3.6.2 How to Determine Whether A Number Gives a Relative Extremum

• First Derivative Test
  

  Let \(c\) be a critical number for function \(f\). Suppose that \(f\) is continous on \((a, b)\) and differentiable on \((a, b)\) except possibly at \(c\), and that \(c\) is the only critical number for \(f\) in \((a, b)\). Then

  • \(f(c)\) is a relative maximum of \(f\) if \(f'(x) > 0\) in the interval \((a, c)\) and \(f'(x) < 0\) in the interval \((c, b)\)
  • \(f(c)\) is a relative minimum of \(f\) if \(f'(x) < 0\) in the interval \((a, c)\) and \(f'(x) > 0\) in the interval \((c, b)\)
• Second Derivatitve Test
  

  Suppose \(f''(x)\) exists on some open interval containing \(c\) except possibly at \(c\) itself, and suppose \(f'(c) = 0\).

  • If \(f''(c) > 0\), then \(f(c)\) is a relative minimum
  • If \(f''(c) < 0\), then \(f(c)\) is a relative maximum
  • If \(f''(c) = 0\) or \(f''(c)\) does not exist, then the test gives no information about extrema, so use the first derivative test. (Remark : so, if possible, always use First Derivative Test.)

3.6.3 How to Find All Relative Extrema of a Function

• Procedure
  
  • Find all critical numbers of function \(f(x)\), and separate the domain into several intervals by these critical numbers
  • At each critical number, use First Derivative Test or Second Derivative Test to determine whether it gives a relative maximum or minimum or no information.
• Example
  

  • Using First Derivative Test

    For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know the critical numbers are \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\). Also, we have already got the following table, so we can directly apply the First Derivative Test.

    \(x\)	\((-\infty\), \(\frac{1 - \sqrt{5}}{2})\)	\((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\)	\((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\)	\((\frac{1 + \sqrt{5}}{2}, \infty)\)
    \(f'(x)\)	\(f'(-10) > 0\)	\(f'(0) < 0\)	\(f'(1) < 0\)	\(f'(10) > 0\)
    \(f(x)\)	increasing	desreasing	decreasing	increasing

    • On \((-\infty\), \(\frac{1 - \sqrt{5}}{2})\), \(f'(x) > 0\) and \(f(x)\) is increasing. On \((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\), \(f'(x) < 0\) and \(f(x)\) is decreasing. Thus \(f(\frac{1 - \sqrt{5}}{2})\) is a relative maximum.
    • On \((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\), \(f'(x) < 0\) and \(f(x)\) is decreasing. On \((\frac{1 - \sqrt{5}}{2}, \infty)\), \(f'(x) > 0\) and \(f(x)\) is increasing. Thus \(f(\frac{1 + \sqrt{5}}{2})\) is a relative minimum.
    • \(f(\frac{1}{2})\) is ignored as it is undefined.

  • Using Second Derivative Test

    For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f'(x) = \frac{2x^2 - 2x - 2}{(2x - 1)^2}\), so the second derivative is

    \(f''(x) = [f'(x)]'\)

    \(= \left( \frac{2x^2 - 2x - 2}{(2x - 1)^2} \right)'\)

    \(= \frac{(2x^2 - 2x - 2)'(2x-1)^2 - (2x^2 - 2x - 2)[(2x-1)^2]'}{(2x - 1)^4}\)

    \(= \frac{(4x - 2)(2x-1)^2 - (2x^2 - 2x - 2)[2\cdot 2 \cdot (2x-1)]}{(2x - 1)^4}\)

    \(= \frac{(4x - 2)(2x-1)^2 - 4(2x^2 - 2x - 2) (2x-1)}{(2x - 1)^4}\)

    \(= \frac{(4x - 2)(2x-1) - 4(2x^2 - 2x - 2)}{(2x - 1)^3}\) [by cancelling \((2x-1)\)]

    \(= \frac{8x^2 - 4x - 4x + 2 - 8x^2 + 8x + 8}{(2x - 1)^3}\)

    \(= \frac{10}{(2x-1)^3}\)

    Thus we have

    • \(f'(\frac{1 - \sqrt{5}}{2}) = 0\) and \(f''(\frac{1 - \sqrt{5}}{2}) < 0\), so \(f(\frac{1 - \sqrt{5}}{2})\) is a relative maximum.
    • \(f'(\frac{1 + \sqrt{5}}{2}) = 0\) and \(f''(\frac{1 + \sqrt{5}}{2}) > 0\), so \(f(\frac{1 + \sqrt{5}}{2})\) is a relative minimum.

3.7.1 Definition

3.7.2 For a Given Interval, How to Determine Whether Function \(f(x)\) Is Concave Upward or Downward?

3.7.3 For a Given Function \(f(x)\), How to Find All Intervals on Which It Is Concave Upward or Downcard?

• Procedure
  
  • Find all values of \(x\) such that \(f''(x) = 0\) or \(f''(x)\) does not exist (these values are called possible inflection points, see next section)
  • Separate the whole domain of \(f(x)\) into several intervals by these numbers
  • Within each interval, choose a value of \(x\), say \(a\), and see whether \(f''(a) > 0, < 0\) or \(= 0\).
  • Decide where \(f(x)\) is concave upward or downward on each interval
    • If \(f''(a) > 0\), then \(f(x)\) is concave upward on that interval
    • If \(f''(a) < 0\), then \(f(x)\) is concave downward on that interval
• Example
  

  For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f''(x) = \frac{10}{(2x - 1)^3}\).

  • Find all points where \(f''(x) = 0\) or \(f''(x)\) is undefined.
    • \(f''(x) = 0\) never holds because the numerator \(10\) can never equal \(0\).
    • \(f''(x)\) is undefined at \(x=\frac{1}{2}\)
    • Therefore, \(\frac{1}{2}\) is the only possible inflection point
  • The domain of the original function \(f(x)\) is \((-\infty,\frac{1}{2}) \cup (\frac{1}{2},\infty)\), so the possible inflection point \(x=\frac{1}{2}\) doesn't separate the domain any more. As a result, the first row of the following table contains only two intervals.
  • Calculate \(f''(x)\) at some special point within each interval.

  • Based on the sign of the second row of the table, we can determine the concavity of the function \(f(x)\) as follows.

    \(x\)	\((-\infty, \frac{1}{2})\)	\((\frac{1}{2}, \infty)\)
    \(f''(x)\)	\(f''(0) < 0\)	\(f''(1) > 0\)
    \(f(x)\)	concave downward	concave upward

3.8.1 Definition

3.8.2 How to Determine Whether a Point is an Inflection Point

3.8.3 How to Find All Inflection Points of a Function

• Procedure
  
  • Find all possible inflection points (candidates for further investigation). A point \(c\) is said to be a possible inflection point if \(f''(c) = 0\) or \(f''(c)\) is not defined.
  • At each possible inflection point, determine by the above method whether it is truly an inflection point.
• Example
  

  For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f''(x) = \frac{10}{(2x - 1)^3}\). \(f''(x)\) is undefined at \(x=\frac{1}{2}\), and \(f''(x) = 0\) can not be achieved for any value of \(x\).

  • Thus \(\frac{1}{2}\) is the only possible inflection points.
  • At \(\frac{1}{2}\), the concavity changes, so \(\frac{1}{2}\) is an inflection point.

3.9.1 Example 1

3.9.2 Example 2

4.1.1 Definition

• Remark
  

  "Anti-Derivative", by its name, is the inverse of derivative.

4.1.2 Property

• Non-Uniqueness of Antiderivative
  

  If the antiderivative of a function exists, then it is not unique.

  For example, both \(F(x) = x^2\) and \(G(x) = x^2 + 1\) are antiderivatives of \(f(x) = 2x\).

• Relationship between Different Antiderivatives
  

  If \(F(x)\) and \(G(x)\) are both antiderivatives of a function \(f(x)\) on an interval, then there is a constant \(C\) such that \[ F(x) - G(x) = C \]

  Remark

  \(F(x) - G(x) = C\) implies \(F(x) = G(x) + C\). That is to say, once a certain antiderivative of \(f(x)\), say \(G(x)\), is found, then all other antiderivates of \(f(x)\) can be obtained as well simply by adding a constant to \(G(x)\).

4.2.1 Definition

4.2.2 Property

4.2.3 Rules

4.2.4 Technique 1 : Rewriting the Expression

• Example 1 : Splitting a Long Numerator
  

  Find \(\int f(x) dx\) if \(f(x) = \frac{2 + 3 x^2 - 4 \sqrt{x}}{x^3}\).

  Rewrite

  \(f(x) = \frac{2 + 3 x^2 - 4 \sqrt{x}}{x^3}\)

  \(= \frac{2}{x^3} + \frac{3}{x} - 4 \frac{\sqrt{x}}{x^3}\)

  \(= \frac{2}{x^3} + \frac{3}{x} - 4 x^{\frac{1}{2}-3}\)

  \(= \frac{2}{x^3} + \frac{3}{x} - 4 x^{\frac{-5}{2}}\)

  Now it is ready to apply power rule and logarithmic rule to find \(\int f(x) dx\).

• Example 2 : Expanding a Product or Power
  

  Find \(\int f(x) dx\) if \(f(x) = (3x^4 - 5)^2\).

  Rewrite

  \(f(x) = (3x^4)^2 + 5^2 - 2 \cdot 3x^4 \cdot 5 = 9x^8 + 25 - 30x^4\)

  Now it is ready to apply power rule to find \(\int f(x) dx\).

• Example 3 : Expanding a Product or Power
  

  Find \(\int f(x) dx\) if \(f(x) = (3x^4 - 5) (6x + 7)\)

  Rewrite

  \(f(x) = (3x^4 - 5) (6x + 7) = 3x^4 \cdot 6x + 3x^4 \cdot 7 - 5 \cdot 6x - 5 \cdot 7 = 18x^5 + 21x^4 - 30x -35\)

  Now it is ready to apply power rule to find \(\int f(x) dx\).

4.2.5 Technique 2 : Integration by Substitution

• Introduction
  

  Inverting the chain rule of differentiation (details are omitted here), we get the Substitution Method of integration.

  Like chain rule, the substitution method is used for dealing with composite functions. By far, the rules and technique we have learned are just suitable for simple functions. As you will see, without the substitution method, it is extremely hard to find the integral of complicated composite functions.

  Considering this is a 1-level course, we won't discuss the formal formula or statement of the substitution method. Instead, let's experience the power of this method through examples. After the last illustrative example, a summary is given.

• Illustrative Examples
  

  • Illustrative Example 1

    Calculate \(\int ( \sqrt{3x^2 - 4} \cdot 6x ) dx\).

    The \(h(x) = \sqrt{3x^2 - 4} \cdot 6x\) is a product of a square root function and a linear function. Obviously, it is hard to directly apply power rule or exponential rule, and rewriting the expression seems not helpful, either.

    If you have a good intuition, you may find \((3x^2 - 4)' = 6x\). Yes, this does help. If we \(\boxed{ \text{let variable } u \text{ be the inner function of the original composite integrand} }\), ie, let

    \[ \boxed{ u = 3x^2 - 4 } \]

    then \[ \frac{du}{dx} = u' = 6x \]

    and thus \(\boxed{u'dx \text{ can be represented by } du}\), ie,

    \[ \boxed{ du = u' dx = 6x dx } \]

    Then by \(\boxed{ \text{replacing the inner function in the original integrand with variable } u}\) and \(\boxed{ \text{ replacing the } u' dx \text{ in the original integral with } du}\), we can \(\boxed{ \text{rewrite the original integral as one only about } u}\)

    \[ \int ( \sqrt{3x^2 - 4} \cdot 6x ) dx = \int (\underbrace{3x^2 - 4}_u)^{\frac{1}{2}} \cdot \underbrace{6x dx}_{du} = \int u^{\frac{1}{2}} du \]

    Look at the RHS, and you will find there is no \(x\) any more. Instead, it becomes an integral about only \(u\), and the form is much simpler than the original one! Let's calculate it using power rule

    \(\int ( \sqrt{3x^2 - 4} \cdot 6x ) dx\) \(= \int u^{\frac{1}{2}} du\) \(= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C\) \(= \frac{2u^{\frac{3}{2}}}{3} + C\)

    Now the answer is in term of \(u\), let's get it back to \(x\) by replacing \(u\) with \(u = 3x^2 - 4\).

    \(\int ( \sqrt{3x^2 - 4} \cdot 6x ) dx\) \(= \frac{2u^{\frac{3}{2}}}{3} + C\) \(= \frac{2(3x^2 - 4)^{\frac{3}{2}}}{3} + C\)

    This is impressive! Just by defining \(u\) as the inner function of the original compostite integrand, we have made things much simpler.

  • Illustrative Example 2

    Calculate \(\int (\sqrt{3x^2 - 4} \cdot 2x) dx\).

    This is a little bit harder than Example 1. Again we let

    \[ u = 3x^2 - 4 \]

    Find \(\frac{du}{dx} = u' = 6x\) and thus

    \[ du = u' dx = 6x dx \]

    As before, we may simply replace the inner function \(3x^2 - 4\) in the original integral with variable \(u\), but this time \(u' dx = 6x dx\) is not in the original integral \(\int ( \sqrt{3x^2 - 4} \cdot 2x ) dx\), so we can't directly replace it with \(du\). In this case, we should let \(u' dx\) appear by \(\boxed{ \text{multiplying the orginal integrand by } u' \text{ and then dividing it by } u'}\). In this example, \(u=6x\), so we have

    \(\int ( \sqrt{3x^2 - 4} \cdot 2x ) dx\)

    \(= \int ( \sqrt{3x^2 - 4} \cdot 2x \cdot \frac{6x}{6x}) dx\)

    \(= \int (\underbrace{3x^2 - 4}_u)^{\frac{1}{2}} \cdot \frac{2x}{6x} \cdot \underbrace{6xdx}_{du}\)

    \(= \frac{1}{3} \int u^{\frac{1}{2}} du\)

    \(= \frac{1}{3} (\frac{2u^{\frac{3}{2}}}{3} + C)\)

    \(= \frac{2u^{\frac{3}{2}}}{9} + C^*\)

    \(= \frac{2(3x^2 - 4)^{\frac{3}{2}}}{9} + C^*\)

    This example tells us that sometimes you need rewrite the orginal integral to let \(u'dx\) appear and then replace it with \(du\). Usually, we do this by multiplying the original integrand by \(u'\) and then dividing it by \(u'\).

  • Illustrative Example 3

    Calculate \(\int ( \sqrt{3x - 4} \cdot 2x ) dx\).

    This example is in turn a little bit harder than Example 2. Again we let

    \[ u = 3x - 4 \]

    Find \(\frac{du}{dx} = u' = 3\) and thus

    and find

    \[ du = u'dx = 3 dx \]

    Again \(u' dx = 3dx\) is not in the original integral \(\int ( \sqrt{3x - 4} \cdot 2x ) dx\) so we multiply the integrand by \(u'=3\) and then divide it by \(u'=3\) to let \(u'dx = 3dx\) appear:

    \(\int ( \sqrt{3x - 4} \cdot 2x ) dx\)

    \(=\int ( \sqrt{3x - 4} \cdot 2x \cdot \frac{3}{3}) dx\)

    \(=\int (\underbrace{3x - 4}_{u})^{\frac{1}{2}} \cdot \frac{2x}{3} \cdot \underbrace{3dx}_{du}\)

    \(=\frac{2}{3} \int (u^{\frac{1}{2}}x) du\)

    Wait, we want to convert the original integral into one only about \(u\), just like those in Example 1 and Example 2, but now we still have \(x\) in the expression after substitution. So let's give up and try another \(u\) or even another method?

    Wait, is it possible to \(\boxed{ \text{replace } x \text{ with a function of } u}\) by chance? Yes, look at our definition of \(u = 3x - 4\), and you get \(x = \frac{u + 4}{3}\), which helps us out:

    \(\int ( \sqrt{3x - 4} \cdot 2x ) dx\)

    \(=\frac{2}{3} \int (u^{\frac{1}{2}} \cdot \underbrace{x}_{\frac{u + 4}{3}} ) du\)

    \(=\frac{2}{3} \int ( u^{\frac{1}{2}} \cdot \frac{u + 4}{3} ) du\)

    Now the last integral is only about \(u\) and there is no \(x\), so we can find the integral

    \(\int ( \sqrt{3x - 4} \cdot 2x ) dx\)

    \(=\frac{2}{3} \int ( u^{\frac{1}{2}} \cdot \frac{u + 4}{3} ) du\)

    \(=\frac{2}{9} \int u^{\frac{1}{2}} \cdot (u + 4) du\)

    \(=\frac{2}{9} \int ( u^{\frac{1}{2}} u + 4u^{\frac{1}{2}} ) du\)

    \(=\frac{2}{9} \int ( u^{\frac{3}{2}} + 4u^{\frac{1}{2}} ) du\)

    \(=\frac{2}{9} (\frac{2u^{\frac{5}{2}}}{5} + 4 \cdot \frac{2u^{\frac{3}{2}}}{3} + C )\)

    \(=\frac{4u^{\frac{5}{2}}}{45} + \frac{16u^{\frac{3}{2}}}{27} + C^*\)

    \(=\frac{4(3x-4)^{\frac{5}{2}}}{45} + \frac{16(3x-4)^{\frac{3}{2}}}{27} + C^*\)

    This example tells us that if the expression after substitution still contains \(x\), try to replace \(x\) with a function of \(u\).

• General Procedure of Substitution Method
  

  Suppose a composite function \(h(x)\) is given and we want to find \(\int h(x) dx\). Then we follow the steps below to apply the substitution method of integration.

  1. Identify the inner function \(g(x)\) and let \(u = g(x)\)
  2. Replace \(g(x)\) with \(u\) in the original integral
  3. Replace \(u'dx\) with \(du\) in the original integral
     1. Find \(u' = g'(x)\)
     2. If \(u'dx\) doesn't appear in the integral yet, then multiply the integrand by \(u'\) and then divide it by \(u'\)
     3. Replace \(u' dx\) with \(du\) in the integral
  4. If now the integral is only about \(u\), then calculate it.
  5. If now the integral still contains \(x\), then replace \(x\) with a function of \(u\). The relationship between \(x\) and \(u\) is determined when we define \(u=g(x)\). Once the integral contains only \(u\), calculate it.
• How to Identify the Inner Function \(g(x)\) and Define \(u\)
  
  • We have already known the power of the substitution method, but for a specific function, how should we identify the inner function \(g(x)\) and then define \(u = g(x)\)?
  • Usually, try to let \(u = g(x)\) be
    • the quantity under a root symbol
    • the base of a power
    • the exponent of a power
    • the denominator of a fraction
• More Examples
  

  • Example 4: \(u=\) base of a power, \(h(x) = (3x-2)^{10}\)

    Theoretically, we may expand the power and then apply the power rule to get \(\int h(x) dx\). However, although theoretically possible, it is not practically recommended to expand a power of order greater than 2. Let's try to apply the substitution method.

    \(h(x)\) is a power function whose base is another function, so let \(u\) be the base part, and we have

    \[ u = 3x - 2 \]

    and thus

    \[ du = u'dx = 3dx \]

    Then we have

    \(\int h(x) dx = \int (3x-2)^{10} dx\)

    \(= \int (3x-2)^{10} \cdot \frac{3}{3}dx\)

    \(= \frac{1}{3} \int (\underbrace{3x-2}_u)^{10} \cdot \underbrace{3dx}_{du}\)

    \(= \frac{1}{3} \int u^{10} du\)

    \(= \frac{1}{3} (\frac{u^{11}}{11} + C)\)

    \(= \frac{u^{11}}{33} + C^*\)

    \(= \frac{(3x-2)^{11}}{33} + C^*\)

  • Example 5: \(u=\) base of a power, \(h(x) = (12x - 8)(3x^2 - 4x + 5)^{100}\)

    Again, it is not wise to expand \(h(x)\). Let's try the substitution method.

    One factor of \(h(x)\) is a power function whose base is another function, so let \(u\) be the base part, and we have

    \[ u = 3x^2 - 4x + 5 \]

    and thus

    \[ du = u'dx = (6x - 4)dx \]

    Then we have

    \(\int h(x) dx = \int (12x - 8)(3x^2 - 4x + 5)^{100} dx\)

    \(=\int 2 (6x - 4)(3x^2 - 4x + 5)^{100} dx\)

    \(=2 \int (\underbrace{3x^2 - 4x + 5}_{u})^{100} \cdot \underbrace{(6x - 4) dx}_{du}\)

    \(=2 \int u^{100} du\)

    \(=2 \cdot \left( \frac{u^{101}}{101} + C \right)\)

    \(=\frac{2 u^{101}}{101} + C^*\)

    \(=\frac{2 (3x^2 - 4x + 5)^{101}}{101} + C^*\)

  • Example 6: \(u=\) base of a power, \(h(x) = \frac{(lnx)^9}{2x}\)

    One factor of \(h(x)\) is a power function whose base is another function, so let \(u\) be the base part, and we have

    \[ u = lnx \]

    and thus

    \[ du = u'dx = \frac{1}{x} dx \]

    Then we have

    \(\int h(x) dx = \int \frac{(lnx)^9}{2x} dx\)

    \(= \frac{1}{2} \int (\underbrace{lnx}_{u})^9 \cdot \underbrace{\frac{1}{x} dx}_{du}\)

    \(= \frac{1}{2} \int u^9 du\)

    \(= \frac{1}{2} \left( \frac{u^{10}}{10} + C \right)\)

    \(= \frac{u^{10}}{20} + C^*\)

    \(= \frac{(lnx)^{10}}{20} + C^*\)

  • Example 7: \(u=\) exponent of a power, \(h(x) = 6 e^{3x-2}\)

    One factor of \(h(x)\) is a exponential function whose exponent is another function, so let \(u\) be the exponent part, and we have

    \[ u = 3x - 2 \]

    and thus

    \[ du = u'dx = 3 dx \]

    Then we have

    \(\int h(x) dx = \int 6 e^{3x-2} dx\)

    \(= 2 \int e^{\overbrace{3x-2}^u} \cdot \underbrace{3 dx}_{du}\)

    \(= 2 \int e^u du\)

    \(= 2 (e^u + C)\)

    \(= 2 e^u + C^*\)

    \(= 2 e^{3x-2} + C^*\)

  • Example 8: \(u=\) exponent of a power, \(h(x) = 6x^2 e^{4x^3 - 1}\)

    One factor of \(h(x)\) is a exponential function whose exponent is another function, so let \(u\) be the exponent part, and we have

    \[ u = 4x^3 - 1 \]

    and thus

    \[ du = u'dx = 12 x^2 dx \]

    Then we have

    \(\int h(x) dx = \int 6x^2 e^{4x^3 - 1} dx\)

    \(= \int e^{4x^3 - 1} \cdot 6x^2 \cdot \frac{2}{2}dx\)

    \(= \frac{1}{2} \int e^{\overbrace{4x^3 - 1}^u} \cdot \underbrace{12x^2 dx}_{du}\)

    \(= \frac{1}{2} \int e^u du\)

    \(= \frac{1}{2} \cdot (e^u + C)\)

    \(= \frac{1}{2} \cdot e^{4x^3 - 1} + C^*\)

  • Example 9: \(u=\) denominator of a fraction, \(h(x) = \frac{2}{3x-2}\)

    \(h(x)\) is a fraction whose denominator is another function, so let \(u\) be the denominator, and we have

    \[ u = 3x - 2 \]

    and thus

    \[ du = u'dx = 3 dx \]

    Then we have

    \(\int h(x) dx = \int \frac{2}{3x-2} dx\)

    \(= \int \frac{2}{3x-2} \cdot \frac{3}{3} dx\)

    \(= \frac{2}{3} \int \frac{1}{\underbrace{3x-2}_u} \cdot \underbrace{3dx}_{du}\)

    \(= \frac{2}{3} \int \frac{1}{u} du\).

    \(= \frac{2}{3} \cdot (ln|u| + C)\)

    \(= \frac{2}{3} \cdot ln|3x-2| + C^*\)

  • Example 10: \(u=\) denominator of a fraction, \(h(x) = \frac{3x}{2x-1}\)

    \(h(x)\) is a fraction whose denominator is another function, so let \(u\) be the denominator, and we have

    \[ u = 2x - 1 \]

    and thus

    \[ du = u'dx = 2 dx \]

    Then we have

    \(\int h(x) dx = \int \frac{3x}{2x-1} dx\)

    \(= \int \frac{3x}{2x-1} \cdot \frac{2}{2} dx\)

    \(= \frac{3}{2} \int \frac{x}{\underbrace{2x-1}_u} \cdot \underbrace{2dx}_{du}\)

    \(= \frac{3}{2} \int \frac{x}{u} du\)

    There is still a \(x\) in the expression after substitution, so we need find the relationship between \(x\) and \(u\) and replace \(x\) with a function of \(u\). Since \(u = 2x-1\), we have \(x = \frac{u+1}{2}\). Therefore,

    \(\int h(x) dx = \int \frac{3x}{2x-1} dx\)

    \(= \frac{3}{2} \int \frac{x}{u} du\)

    \(= \frac{3}{2} \int \frac{u+1}{2u} du\)

    \(= \frac{3}{4} \int \frac{u+1}{u} du\)

    \(= \frac{3}{4} \int \left( 1 + \frac{1}{u} \right) du\)

    \(= \frac{3}{4} (u + ln|u| + C)\)

    \(= \frac{3}{4} (2x - 1 + ln|2x - 1| + C)\)

    \(= \frac{3x}{2} - \frac{3}{4} + \frac{3ln|2x - 1|}{4} + C^*\)

  • Example 11: \(u=\) function under root symbol, \(h(x) = \frac{3x}{\sqrt{2x-1}}\)

    \(h(x)\) is a fraction whose denominator is another function. In turn, the denominator is a square root of some other function. In this case, we let \(u\) be the part under the root symbol, and we have

    \[ u = 2x - 1 \]

    and thus

    \[ du = u'dx = 2 dx \]

    Then we have

    \(\int h(x) dx = \int \frac{3x}{\sqrt{2x-1}} dx\)

    \(= \int \frac{3x}{(2x-1)^{\frac{1}{2}}} \cdot \frac{2}{2} dx\)

    \(= \frac{3}{2} \int \frac{x}{(\underbrace{2x - 1}_u)^{\frac{1}{2}}} \cdot \underbrace{2dx}_{du}\)

    \(= \frac{3}{2} \int \frac{x}{u^{\frac{1}{2}}} du\)

    Again, there is an \(x\) in the expression after substitution. Since \(u = 2x-1\), we have \(x = \frac{u+1}{2}\). Therefore,

    \(\int h(x) dx = \int \frac{3x}{\sqrt{2x-1}} dx\)

    \(= \frac{3}{2} \int \frac{x}{u^{\frac{1}{2}}} du\)

    \(= \frac{3}{2} \int \frac{u+1}{2u^{\frac{1}{2}}} du\)

    \(= \frac{3}{4} \int \frac{u+1}{u^{\frac{1}{2}}} du\)

    \(= \frac{3}{4} \int (u+1)u^{\frac{-1}{2}} du\)

    \(= \frac{3}{4} \int (u \cdot u^{\frac{-1}{2}} + u^{\frac{-1}{2}} )du\)

    \(= \frac{3}{4} \int (u^{\frac{1}{2}} + u^{\frac{-1}{2}} )du\)

    \(= \frac{3}{4} (\frac{2u^{\frac{3}{2}}}{3} + 2u^{\frac{1}{2}} + C)\)

    \(= \frac{u^{\frac{3}{2}}}{2} + \frac{3u^{\frac{1}{2}}}{2} + C^*\)

    \(= \frac{(2x - 1)^{\frac{3}{2}}}{2} + \frac{3(2x-1)^{\frac{1}{2}}}{2} + C^*\)

4.2.6 Technique 3 : Integration by Parts

• Formula
  

  Inverting the product rule of differentiation (details are omitted here), we get the method of Integration by Parts.

  If \(u(x)\) and \(v(x)\) are differentiable functions, then \[ \int (u \cdot v') dx = u \cdot v - \int (u' \cdot v) dx \]

• Remark
  

  For a given integral whose integrand is a product of two factors, if we want to apply integration by parts, how to determine which is \(u\) and which is \(v\)?

  The reason why we apply a method is that the method makes the problem simpler and easier. Hence we should choose \(u\) and \(v\) in such a way that \(\int (u' \cdot v) dx\) is easy to calculate. See the following examples.

• Examples
  

  • Example 1 : \(\int (ax+b)e^{cx+d} dx\), Type 1 Problem

    Find \(\int (2x + 4) e^{3x} dx\).

    This indefinite integral can't be calculated by rules or techniques that we have learned in previous sections. Since the integrand is a product of two factors, we may try integration by parts.

    Should we let

    • \(u = 2x + 4\), \(v' = e^{3x}\), or
    • \(u = e^{3x}\), \(v' = 2x + 4\)?

    As mentioned before, we should define \(u\) and \(v\) in such a way that \(\int u'v dx\) is easy to calculate. In this example, \((2x+4)'=2\) and \((e^{3x})' = 3e^{3x}\). It may be wise to let \(u = 2x + 4\) because then \(u' = 2\) is very simple. Accordingly, \(v\) should be defined in such a way that \(v' = e^{3x}\), which means \(v = \int e^{3x} dx = \frac{e^{3x}}{3} + C\). Without loss of generality, we let \(C=0\) and thus \(v = \frac{e^{3x}}{3}\).

    By the formula of integration by parts, we have \(\int u v' dx = u v - \int u' v dx\) and thus

    \(\int [(2x + 4) \cdot e^{3x}] dx = (2x + 4) \cdot \frac{e^{3x}}{3} - \int 2 \cdot \frac{e^{3x}}{3} dx\)

    \(= (2x + 4) \cdot \frac{e^{3x}}{3} - \frac{2}{3} \int e^{3x} dx\)

    \(= (2x + 4) \cdot \frac{e^{3x}}{3} - \frac{2}{3} \cdot \left( \frac{e^{3x}}{3} + C \right)\)

    \(= (2x + 4) \cdot \frac{e^{3x}}{3} - \frac{2e^{3x}}{9} + C^*\)

  • Example 2 : \(\int (a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + b)e^{cx+d} dx\), Generalized Type 1 Problem, Iterated Application of Integration by Parts

    Find \(\int (x^2 + 5) e^{3x + 4} dx\).

    This integral \(\int (x^2 + 5) e^{3x + 4} dx\) doesn't fit in the simple Type 1 integration-by-parts problem \(\int (ax+b)e^{cx+d} dx\) in Example 1, because \(x^2 + 5\) doesn't fit in \((ax + b)\). However, let's still apply integration by parts to see what we get. Similar to Example 1, we let \(u=x^2 + 5\) and \(v'=e^{3x + 4}\). Then \(u'=2x\) and \(v = \int e^{3x + 4} dx = \frac{e^{3x + 4}}{3} + C\). Without loss of generality, let \(C=0\) and thus \(v=\frac{e^{3x + 4}}{3}\).

    Applying integration by parts \(\int uv' dx = uv - \int u'v dx\), we have

    \(\int (x^2 + 5) e^{3x+4} dx\)

    \(= (x^2 + 5) \cdot \frac{e^{3x+4}}{3} - \int 2x \cdot \frac{e^{3x+4}}{3} dx\)

    \(= (x^2 + 5) \cdot \frac{e^{3x+4}}{3} - \frac{2}{3} \int x e^{3x+4} dx\)

    Now we need find \(\int x e^{3x+4} dx\), which is in form of \((ax+b)e^{cx+d}\), similar to Example 1, so we use integration by parts again. Note this is the second time we apply integration by parts for this problem. Since symbols \(u\) and \(v\) are already used, to avoid confusion, this time we let \(s=x\) and \(t' = e^{3x+4}\). Then \(s' = 1\) and \(t = \int e^{3x+4} dx = \frac{e^{3x+4}}{3} + C\). Without loss of generality, let \(C=0\) and thus \(t=\frac{e^{3x+4}}{3}\).

    Applying integration by parts \(\int st' dx = st - \int s't dx\), we have

    \(\int x e^{3x+4} dx\)

    \(= x \cdot \frac{e^{3x+4}}{3} - \int 1 \cdot \frac{e^{3x+4}}{3} dx\)

    \(= x \cdot \frac{e^{3x+4}}{3} - \frac{1}{3} \int e^{3x+4} dx\)

    \(= x \cdot \frac{e^{3x+4}}{3} - \frac{1}{3} \cdot \frac{e^{3x+4}}{3} + C\)

    \(= x \cdot \frac{e^{3x+4}}{3} - \frac{e^{3x+4}}{9} + C\)

    Finally, by putting the above result into the previous one, we have

    \(\int (x^2 + 5) e^{3x+4} dx\)

    \(= (x^2 + 5) \cdot \frac{e^{3x+4}}{3} - \frac{2}{3} \int x e^{3x+4} dx\)

    \(= (x^2 + 5) \cdot \frac{e^{3x+4}}{3} - \frac{2}{3} \cdot (x \cdot \frac{e^{3x+4}}{3} - \frac{e^{3x+4}}{9} + C)\)

    \(= (x^2 + 5) \cdot \frac{e^{3x+4}}{3} - \frac{2}{3} \cdot (x \cdot \frac{e^{3x+4}}{3} - \frac{e^{3x+4}}{9}) + C^*\)

    which is just the answer.

    Here is nothing mysterious. We just applied integration by parts for twice. That's it. The order of the polynomial \(x^2 + 5\) in the original integrand is \(2\), so we apply the integration by parts for \(2\) times in total. Generally, for \(\int (a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + b)e^{cx+d} dx\), we apply the integration by parts for \(n\) times. For each time, we let \(u\) be the polynomial and let \(v'\) be \(e^{cx+d}\). Apparently, Example 1 is a special case of this example corresponding to \(n=1\), which applies integration by parts for only once.

    Caution : when applying integration by parts for more than once, we should use different symbols for the two variables in each round to avoid confusion. For example, in the first round, we may use \(u\) and \(v\), while in the second round we should use \(s\) and \(t\), etc.

  • Example 3 : \(\int x^a ln(x^b) dx\), Type 2 Problem

    Find \(\int x^7 \cdot ln(x^5) dx\).

    This indefinite integral can't be calculated by rules or techniques that we have learned in previous sections. Since the integrand is a product of two factors, we may try integration by parts.

    Should we let

    • \(u = x^7\), \(v' = ln(x^5)\) or
    • \(u = ln(x^5)\), \(v' = x^7\)?

    As mentioned before, we should define \(u\) and \(v\) in such a way that \(\int u'v dx\) is simple. In this example, \((x^7) = 7 x^6\) and \((ln(x^5))' = \frac{1}{x^5} \cdot 5x^4 = \frac{5}{x}\). It may be wise to let \(u = ln(x^5)\) because then \(u' = \frac{5}{x}\) is simple. Accordingly, \(v\) should be defined in such a way that \(v' = x^7\), which means \(v = \int x^7 dx = \frac{x^8}{8} + C\). Without loss of generality, we let \(C = 0\) and thus \(v = \frac{x^8}{8}\).

    By the formula of integration by parts, we have \(\int u v' dx = u \cdot v - \int u' v dx\) and thus

    \(\int ln(x^5) \cdot x^7 dx = ln(x^5) \cdot \frac{x^8}{8} - \int \frac{5}{x} \cdot \frac{x^8}{8} dx\)

    \(= ln(x^5) \cdot \frac{x^8}{8} - \frac{5}{8} \cdot \int x^7 dx\)

    \(= ln(x^5) \cdot \frac{x^8}{8} - \frac{5}{8} \cdot \left( \frac{x^8}{8} + C \right)\)

    \(= ln(x^5) \cdot \frac{x^8}{8} - \frac{5x^8}{64} + C^*\)

  • Example 4 : \(\int ln(x) dx\), Type 2 Problem, Special Case

    Find \(\int ln(x) dx\).

    This is a special case of Type 2 Problem corresponding to \(a = 0\) and \(b = 1\). We can rewrite the original problem as \(\int ln(x) dx = \int x^0 ln(x) dx = \int 1 \cdot ln(x) dx\).

    Similar to Example 3, let \(u = ln(x)\) and \(v' = 1\). Then \(v = \int v' dx = \int 1 dx = x + C\). Without loss of generality, let \(C=0\) and thus \(v = x\).

    By the formula of integration by parts \(\int u v' dx = u \cdot v - \int u' v dx\), we have

    \(\int ln(x) dx\)

    \(= \int ln(x) \cdot 1 dx = ln(x) \cdot x - \int (lnx)'x dx\)

    \(= lnx \cdot x - \int \frac{1}{x} \cdot x dx\)

    \(= lnx \cdot x - \int 1 dx\)

    \(= lnx \cdot x - x + C\)

• Summary
  

  In this course (not true in general), there are mainly 2 types of problems suitable for integration by parts:

  • Type 1 : \(\int (a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + b)e^{cx+d} dx\) where \(a_1, a_2, \dots, a_n, b, c\) and \(d\) are constants. See Example 2. When \(n=1\), the integral becomes \(\int (ax + b)e^{cx+d} dx\). See Example 1.
  • Type 2 : \(\int x^a ln(x^b) dx\) where \(a\) and \(b\) are constant. See Example 3 and Example 4.

  According to the examples,

  • If the integrand is a product in form of \((a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + b)e^{cx+d}\), ie, Type 1, then use integraion by parts for \(n\) times. For the first application, let \(u = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + b\) and \(v'=e^{cx+d}\). The similar strategy for the second and further applications. Especially, if the integrand is in form of \((ax + b)e^{cx+d}\), then one application of integration by part is enough, by letting \(u=ax+b\) and \(v'=e^{cx+d}\). When applying integration by parts for more than once, we should use different symbols for the two variables in each round to avoid confusion. For example, in the first round, we may use \(u\) and \(v\), while in the second round we should use \(s\) and \(t\), etc.
  • If the integrand is a product in form of \(x^a ln(x^b)\), ie, Type 2, then use integration by parts by letting \(u=ln(x^b)\) and \(v'=x^a\).
  • Otherwise, try other methods.
    • For example, if the integrand is a product of two polynomials, like \(\int (ax+b)(cx+d) dx\), then it is of neither Type 1 nor Type 2. We solve this problem by rewriting (expanding) the product \(\int (ax+b)(cx+d) dx\) \(=\int (ac\cdot x^2 + ad\cdot x + bc\cdot x + bd) dx\) \(= ac\int x^2 dx + ad\int x dx + bc\int x dx + bd\int 1 dx\)
    • For example, if the integrand is \(xe^{x^2}\), then it is a product of neither Type 1 nor Type 2. We solve this problem using integration by substitution. Let \(u = x^2\) and thus \(du = u' dx = 2x dx\). then we have \(\int xe^{x^2} dx\) \(= \frac{1}{2} \int e^{x^2} 2x dx\) \(= \frac{1}{2} \int e^u du\) \(= \frac{1}{2} \cdot e^u + C\) \(= \frac{1}{2} \cdot e^{x^2} + C\).

4.2.7 Discussion : Integration by Substitution and Integration by Parts

• Comparison
  

  As we have seen, both Integration by Substitution and Integration by Parts are used for integrands in form of product of two factors, so when should we use Integration by Substitution and when should we use Integration by Parts?

  Recall that

  • For Integration by Substitution (inversion of Chain Rule of differentiation),
    • the formula is \(\int f(u(x)) u'(x) dx = \int f(u) du\)
    • we define only one new variable \(u = u(x)\)
    • \(u'\) should appear in the original integrand
    • we replace \(u(x)\) in the original integrand with variable \(u\)
    • we replace \(u'dx\) in the original integrand with \(du\)
    • we convert the original integral into an integral \(\int f(u) du\) about variable \(u\) only
    • the new integral \(\int f(u) du\) is with respcet to the new variable \(u\) only, so there shouldn't be any \(x\) in the integrand \(f(u)\)
  • For Integration by Parts (inversion of Production Rule of differentiation),
    • the formula is \(\int u(x)v'(x) dx = u(x)v(x) - \int u'(x)v(x) dx\)
    • we define two new variables \(u\) and \(v\)
    • \(v'\) should appear in the integrand
    • we don't replace anything in the original integrand (We just figure out which factor is \(u\) and which factor is \(v'\). We don't really replace the factors with symbols \(u\) and \(v'\). That is, No Substitution.)
    • we convert the original integral into an expression which contains \(\int (u'v) dx\)
    • the new integral \(\int (u'v) dx\) is with respect to the original variable \(x\), so the integrand \(u'v\) should be given in term of \(x\) and there is no symbol like \(u,v,u'\) or \(v'\)

  Clearly, integration by substitution and integration by parts are different in nature. As a result, they have different scopes of applications. In most cases for this course, only one of them works for each step. Please see the following examples.

• Examples
  

  • Example 1 : Integration by Parts

    Calculate \(\int x e^{2x} dx\).

    If we use integration by substitution by letting \(u = 2x\) (so \(x = \frac{u}{2}\)) and thus \(u'=2\), then the original integral becomes

    \(\int x e^{2x} dx\)

    \(= \int x e^{2x} \cdot \frac{2}{2} dx\)

    \(= \frac{1}{2} \int x e^{2x} \cdot 2 dx\)

    \(= \frac{1}{2} \int x e^u du\)

    \(= \frac{1}{2} \int \frac{u}{2} \cdot e^u du\)

    \(= \frac{1}{4} \int u e^u du\)

    The integrand is still a product of two factors, which is not substantially simpler than the original integral. Therefore, integration by substitution doesn't work for this problem.

    The integrand here is a product in form of \((ax+b)e^{cx+d}\), ie, Type 1 of the integration-by-parts problems, so Integration by Parts should work.

    In this example, \(x'=1\) and \((e^{2x})' = 2 e^{2x}\). It may be wise to let \(u=x\) because then \(x'\) is simpler than \((e^{2x})'\). Accordingly, \(v\) should be defined in such at way that \(v'=e^{2x}\), which means \(v = \int e^{2x} dx = \frac{e^{2x}}{2} + C\). Without loss of generality, let \(C=0\) and thus \(v = \frac{e^{2x}}{2}\).

    By the formula of integration by parts, we have \(\int u v' dx = u v - \int u' v dx\) and thus

    \(\int x e^{2x} dx = x \cdot \frac{e^{2x}}{2} - \int (1 \cdot \frac{e^{2x}}{2}) dx\)

    \(= x \cdot \frac{e^{2x}}{2} - \frac{1}{2} \int e^{2x} dx\)

    \(= x \cdot \frac{e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + C\)

    \(= e^{2x} \cdot \frac{x}{2} - e^{2x} \cdot \frac{1}{4} + C\)

    \(= e^{2x} \cdot \frac{2x - 1}{4} + C\)

  • Example 2 : Integration by Substitution

    Calculate \(\int x e^{x^2} dx\).

    The integrand here is a product of neither Type 1 nor Type 2 of the integration-by-parts problems, so integration by parts may not apply. However, integration by substitution works for this example.

    Let \(u = x^2\). Then \(du = u'dx = 2x dx\).

    The original integral becomes

    \(\int x e^{x^2} dx\)

    \(= \int e^{x^2} \cdot x \cdot \frac{2}{2} dx\)

    \(= \frac{1}{2} \int e^{x^2} \cdot 2x dx\)

    \(= \frac{1}{2} \int e^u du\)

    \(= \frac{1}{2} (e^u + C)\)

    \(= \frac{1}{2} (e^{x^2} + C)\)

    \(= \frac{e^{x^2}}{2} + C^*\)

    Remark : Example 1 is \(\int x e^{2x} dx\) and Example 2 is \(\int x e^{x^2} dx\). The two integrals look similar, but need different techniques to calculate.

  • Example 3 : Using both Integration by Substitution and Integration by Parts

    Find \(\int 6x^7 \cdot ln(3x^8 + 5) dx\).

    The Type 2 integration-by-parts is \(\int x^a ln(x^b) dx\), so this example \(\int 6x^7 \cdot ln(3x^8 + 3) dx\) doesn't fit in the Type 2 integration-by-parts problem.

    Since integration by parts doesn't work, let's try integration by substitution. Let \(u = 3x^8 + 5\). Then we have \(u' = 24 x^7\) and thus \(du = u' dx = 24 x^7 dx\). Therefore,

    \(\int 6x^7 \cdot ln(3x^8 + 5) dx\)

    \(= \int 6x^7 \cdot ln(3x^8 + 5) \frac{24}{24}dx\)

    \(= \frac{6}{24} \int ln(3x^8 + 5) \cdot 24 x^7 dx\)

    \(= \frac{1}{4} \int ln(u) du\)

    Now we need find \(\int ln(u) du\), which fits in the Type 2 integration-by-parts problem. To apply integration by parts, we need define two new variables like "\(u\)" and "\(v\)". However, we have already used symbol "\(u\)" in the previous substitution and now \(u\) is the "original variable" in the integral \(\int ln(u) du\). Thus, to avoid confusion, let's use symbols \(s\) and \(t\) this time for the integration by parts. Define \(s = ln(u)\) and \(t' = 1\). By \(t' = 1\), we get \(t = \int t' du = \int 1 du = u + C\). Without loss of generality, let \(C = 0\) and thus \(t = u\). For this problem, the formula for integration by parts is \(\int s t' du = st - \int s't du\), and thus

    \(\int ln(u) du = \int ln(u) \cdot 1 du = ln(u) \cdot u - \int \frac{1}{u} \cdot u du\)

    \(= ln(u) \cdot u - \int 1 du\)

    \(= ln(u) \cdot u - u + C\).

    Putting the result back into the previous expression, we get

    \(\int 6x^7 \cdot ln(3x^8 + 5) dx\)

    \(= \frac{1}{4} \int ln(u) du\)

    \(= \frac{1}{4} (ln(u) \cdot u - u + C)\)

    Replacing \(u\) with \(3x^8 + 5\), we get

    \(\int 6x^7 \cdot ln(3x^8 + 5) dx\)

    \(= \frac{1}{4} (ln(u) \cdot u - u + C)\)

    \(= \frac{1}{4} (ln(3x^8 + 5) \cdot (3x^8 + 5) - (3x^8 + 5) + C)\)

    \(= \frac{ln(3x^8 + 5) \cdot (3x^8 + 5)}{4} - \frac{3x^8 + 5}{4} + C^*\)

  • Example 4 : Using either Integration by Substitution or Integration by Parts

    Find \(\int \frac{1}{x} \cdot ln(x) dx\).

    • Method 1 : Integration by Substitution

      Let \(u = ln(x)\). Then \(du = u' dx = \frac{1}{x} dx\) and thus

      \(\int \frac{1}{x} \cdot ln(x) dx\)

      \(= \int ln(x) \cdot \frac{1}{x} dx\)

      \(= \int u du\)

      \(= \frac{u^2}{2} + C\)

      \(= \frac{[ln(x)]^2}{2} + C\)

    • Method 2 : Integration by Parts

      Let \(u = ln(x)\) and \(v' = \frac{1}{x}\). Then \(u' = \frac{1}{x}\) and \(v = \int v' dx = \int \frac{1}{x} dx = ln|x| + C\). The original integral \(\int \frac{1}{x} \cdot ln(x) dx\) is defined only for \(x > 0\), so \(v = ln|x| + C = ln(x) + C\). Therefore, by the formula for integration by parts \(\int uv' dx = uv - \int u'v dx\), we have

      \(\int ln(x) \cdot \frac{1}{x} dx = ln(x) \cdot (ln(x) + C) - \int \frac{1}{x} \cdot (ln(x) + C) dx\)

      \(\implies \int ln(x) \cdot \frac{1}{x} dx = ln(x) \cdot ln(x) + C \cdot ln(x) - \int \frac{1}{x} \cdot ln(x) dx - C \int \frac{1}{x} dx\)

      Adding \(\int ln(x) \cdot \frac{1}{x} dx\) to both sides, we get

      \(2 \int ln(x) \cdot \frac{1}{x} dx = ln(x) \cdot ln(x) + C \cdot ln(x) - C \int \frac{1}{x} dx\)

      \(\implies 2 \int ln(x) \cdot \frac{1}{x} dx = [ln(x)]^2 + C \cdot ln(x) - C (ln(x) + C_1)\)

      \(\implies 2 \int ln(x) \cdot \frac{1}{x} dx = [ln(x)]^2 + C \cdot ln(x) - C \cdot ln(x) - C \cdot C_1\)

      \(\implies 2 \int ln(x) \cdot \frac{1}{x} dx = [ln(x)]^2 - C \cdot C_1\)

      \(\implies \int ln(x) \cdot \frac{1}{x} dx = \frac{[ln(x)]^2}{x} - \frac{C \cdot C_1}{2} = \frac{[ln(x)]^2}{x} + C^*\)

      Which is the same as the answer obtained by integration by substitution.

4.3.1 Introduction

4.3.2 Definition

• Remark
  
  • "\(dx\)" means that the integral is with respect to variable \(x\), ie, we regard \(x\) as the variable of interest.
  • \(a\) and \(b\) are called lower limit and upper limit of the definite integral, respectively.
  • In most cases, it is quite hard to calculate a definite integral by definition. Fortunately, the Fundamental Theorem of Calculus can help us convert the problem of definite integral into a problem of indefinite integral.

4.3.3 Properties

• Example
  

  Suppose \(\int_7^1 f(x) dx = 5\) and \(\int_3^7 f(x) dx = 2\). Find \(\int_1^3 (-3) \cdot f(x) dx\).

  We know \(5 = \int_7^1 f(x) dx = - \int_1^7 f(x) dx\), so \(\int_1^7 f(x) dx = -5\).

  Also, \(\int_1^7 f(x) dx = \int_1^3 f(x) dx + \int_3^7 f(x) dx\), so \(\int_1^3 f(x) dx = \int_1^7 f(x) dx - \int_3^7 f(x) dx\).

  Therefore,

  \(\int_1^3 (-3) \cdot f(x) dx\)

  \(= -3 \int_1^3 f(x) dx\)

  \(= -3 \cdot (\int_1^7 f(x) dx - \int_3^7 f(x) dx)\)

  \(= -3 \cdot (-5 - 2)\)

  \(= -3 \cdot (-7)\)

  \(= 21\)

4.4.1 Statement of The Theorem

• Remark
  
  • The theorem tells us that we can convert a problem about definite integral into one about indefinite integral.
  • Without loss of generality, we may let \(F\) be the antiderivative of \(f\) in which the constant term \(C=0\).
  • The requirement "\(f\) be continuous on the interval \([a, b]\)" is necessary.

4.4.2 Examples

4.5.1 Definitions

4.5.2 Useful Limits About Infinity

4.5.3 Examples

• Example 1
  

  Find \(\int_{-\infty}^{0} e^{2x - 1} dx\).

  By exponential rule integration, we know \(\int e^{2x-1} dx = \frac{e^{2x-1}}{2} + C\).

  By FTC, we have \(\int_{a}^{0} e^{2x-1} dx\) \(= \frac{e^{2x-1}}{2} \big|_{a}^{0}\) \(= \frac{e^{2\cdot 0 - 1}}{2} - \frac{e^{2\cdot a - 1}}{2}\) \(= \frac{e^{-1}}{2} - \frac{e^{2a - 1}}{2}\)

  By the formula,

  \(\int_{-\infty}^{0} e^{2x-1} dx\)

  \(= \lim\limits_{a\to -\infty} \int_{a}^{0} e^{2x-1} dx\)

  \(= \lim\limits_{a\to -\infty} \left( \frac{e^{-1}}{2} - \frac{e^{2a-1}}{2} \right)\)

  \(= \frac{e^{-1}}{2} - \frac{ \lim\limits_{a\to -\infty} e^{2a-1}}{2}\)

  \(= \frac{e^{-1}}{2} - \frac{0}{2}\)

  \(= \frac{e^{-1}}{2}\)

  which is a finite number, so the improper integral is convergent.

• Example 2
  

  Find \(\int_{0}^{\infty} e^{2x - 1} dx\).

  By Example 1, we know \(\int_{0}^{b} e^{2x-1} dx\) \(= \frac{e^{2x-1}}{2} \big|_{0}^{b}\) \(= \frac{e^{2\cdot b - 1}}{2} - \frac{e^{2\cdot 0 - 1}}{2}\) \(= \frac{e^{2b-1}}{2} - \frac{e^{- 1}}{2}\)

  By the formula,

  \(\int_{0}^{\infty} e^{2x-1} dx\)

  \(= \lim\limits_{b\to \infty} \int_{0}^{b} e^{2x-1} dx\)

  \(= \lim\limits_{b\to \infty} \left( \frac{e^{2b-1}}{2} - \frac{e^{-1}}{2} \right)\)

  \(= \infty - \frac{e^{-1}}{2}\)

  \(= \infty\)

  so the improper integral is divergent.

• Example 3
  

  Find \(\int_{-\infty}^{1} \frac{2x}{x^2 + 1} dx\).

  We need first find the indefinite integral \(\int \frac{2x}{x^2 + 1} dx\).

  Let \(u=x^2+1\). Then \(u'=(x^2+1)=2x\) and \(du=u'dx = 2xdx\). Thus

  \(\int \frac{2x}{x^2 + 1} dx\)

  \(= \int \frac{1}{x^2 + 1} 2xdx\)

  \(= \int \frac{1}{u} du\)

  \(= ln|u| + C\)

  \(= ln|x^2+1| + C\)

  so the definite integral

  \(\int_a^1 \frac{2x}{x^2 + 1} dx\)

  \(= ln|x^2+1| \big|_a^1\)

  \(= ln|1^2+1| - ln|a^2+1|\)

  \(= ln|2| - ln|a^2+1|\)

  \(= ln2 - ln(a^2 + 1)\)

  \(= ln\left( \frac{2}{a^2 + 1} \right)\)

  By the formula,

  \(\int_{-\infty}^{1} \frac{2x}{x^2 + 1} dx\) \(= \lim\limits_{a\to -\infty} ln\left( \frac{2}{a^2 + 1} \right)\) \(= ln(0^+)\) \(= -\infty\)

  so the improper integral is divergent.

• Example 4
  

  Find \(\int_{1}^{\infty} \frac{2x}{x^2 + 1} dx\).

  Similar to Example 3, we know the definite integral

  \(\int_1^b \frac{2x}{x^2 + 1} dx\)

  \(= ln|x^2+1| \big|_1^b\)

  \(= ln|b^2+1| - ln|1^2+1|\)

  \(= ln|b^2+1| - ln|2|\)

  \(= ln(b^2+1) - ln2\)

  \(= ln\left( \frac{b^2+1}{2} \right)\)

  By the formula,

  \(\int_{1}^{\infty} \frac{2x}{x^2 + 1} dx\) \(= \lim\limits_{b\to \infty} ln\left( \frac{b^2+1}{2} \right)\) \(=ln(\infty)\) \(=\infty\)

  so the improper integral is divergent.

• Example 5
  

  Find \(\int_{-\infty}^{\infty} \frac{2x}{(x^2 + 1)^2} dx\).

  We need first find the indefinite integral \(\int \frac{2x}{(x^2 + 1)^2} dx\).

  Using integration by substitution, let \(u = x^2 + 1\), then \(du= u' dx = (x^2 + 1)' dx = 2x dx\), and

  \(\int \frac{2x}{(x^2 + 1)^2} dx\)

  \(= \int \frac{1}{(x^2 + 1)^2} 2x dx\)

  \(= \int \frac{1}{u^2} du\)

  \(= \int u^{-2} du\)

  \(= - u^{-1} + C\)

  \(= - \frac{1}{u} + C\)

  \(= - \frac{1}{x^2 + 1} + C\)

  By FTC, we have \(\int_{a}^{0} \frac{2x}{(x^2 + 1)^2} dx\) \(= - \frac{1}{x^2 + 1} \big|_{a}^{0}\) \(= \left( - \frac{1}{0^2 + 1} \right) - \left( - \frac{1}{a^2 + 1} \right)\) \(= - 1 + \frac{1}{a^2 + 1}\)

  Similarly, we have \(\int_{0}^{b} \frac{2x}{(x^2 + 1)^2} dx\) \(= - \frac{1}{x^2 + 1} \big|_{0}^{b}\) \(= \left( - \frac{1}{b^2 + 1} \right) - \left( - \frac{1}{0^2 + 1} \right)\) \(= - \frac{1}{b^2 + 1} + 1\)

  By the formula,

  \(\int_{-\infty}^{\infty} \frac{2x}{(x^2 + 1)^2} dx\)

  \(= \lim\limits_{a\to -\infty} \int_{a}^{0} \frac{2x}{(x^2 + 1)^2} dx + \lim\limits_{b\to \infty} \int_{0}^{b} \frac{2x}{(x^2 + 1)^2} dx\)

  \(= \lim\limits_{a\to -\infty} \left( - 1 + \frac{1}{a^2 + 1} \right) + \lim\limits_{b\to \infty} \left( - \frac{1}{b^2 + 1} + 1 \right)\)

  \(= (- 1 + 0) + ( 0 + 1 )\)

  \(= 0\)

  which is a finite number, so the improper integral is convergent.

4.6.1 Formulae

4.6.2 Procedure

4.6.3 Examples

• Example 1 : Endpoints of the Region Are Given
  

  Find the area of the region bounded by the curves of \(f(x) = 3x^2 + 1\), the \(x\)-axis, \(x=3\) and \(x=5\).

  Let's rephrase the question : find the area of the region between curves of \(f(x) = 3x^2 + 1\) and \(g(x) = 0\) (\(x\)-axis) on interval \([3, 5]\).

  First, we need find all subintervals of \([3, 5]\) on which \(f(x) \ge g(x)\) or \(f(x) \le g(x)\).

  • Find all points where \(f(x) = g(x)\). Let \(f(x) = g(x)\), ie, \(3x^2 + 1 = 0\), and then we see there is no solution to this equation. Hence we don't separate the interval \([3, 5]\) any more.
  • Within \([3, 5]\), we pick a test number \(d = 4\). Then \(f(4) = 3 \cdot 4^2 + 1 = 49\), \(g(4) = 0\), \(f(4) > g(4)\) and thus \(f(x) > g(x)\) on \([3, 5]\).

  Therefore, the area between \(f(x)\) and \(g(x)\) on interval \([3,5]\) equals the definite integral

  \(\int_3^5 [ f(x) - g(x) ] dx\)

  \(= \int_3^5 [ 3x^2 + 1 - 0 ] dx\)

  \(= \int_3^5 (3x^2 + 1) dx\)

  \(= (x^3 + x) \big|_3^5\)

  \(= (5^3 + 5) - (3^3 + 3)\)

  \(= 100\)

• Example 2 : Endpoints of the Region Are Given
  

  Find the area of the region bounded by the curves of \(f(x) = x^2 - 1\), the \(x\)-axis, \(x= -2\) and \(x=3\).

  Let's rephrase the question : find the area of the region between curves of \(f(x) = x^2 - 1\) and \(g(x) = 0\) (\(x\)-axis) on interval \([-2, 3]\).

  First, we need find all subintervals of \([-2, 3]\) on which \(f(x) \ge g(x)\) or \(f(x) \le g(x)\).

  • Find all points where \(f(x) = g(x)\). Let \(f(x) = g(x)\), ie, \(x^2 - 1 = 0\), and then we have \(x = 1\) or \(x = -1\).
  • Separate the interval \([-2, 3]\) by \(-1\) and \(1\), so we get subintervals \((-2, -1), (-1, 1)\) and \((1, 3)\).
  • Within \((-2, -1)\), we pick a test number \(d = -1.5\). Then \(f(-1.5) = (-1.5)^2 - 1 = 1.25\), \(g(-1.5) = 0\), \(f(-1.5) > g(-1.5)\) and thus \(f(x) > g(x)\) on \([-2, -1]\).
  • Within \((-1, 1)\), we pick a test number \(d = 0\). Then \(f(0) = 0^2 - 1 = -1\), \(g(0) = 0\), \(f(0) < g(0)\) and thus \(f(x) < g(x)\) on \((-1, 1)\).
  • Within \((1, 3)\), we pick a test number \(d = 2\). Then \(f(2) = 2^2 - 1 = 3\), \(g(2) = 0\), \(f(2) > g(2)\) and thus \(f(x) > g(x)\) on \((-2, -1)\).

  Then the areas of the regions on subintervals are

  \(A_1 = \int_{-2}^{-1} [ f(x) - g(x) ] dx = \int_{-2}^{-1} [ (x^2 - 1) - 0 ] dx = \left( \frac{x^3}{3} - x \right) \big|_{-2}^{-1} = \frac{4}{3}\)

  \(A_2 = \int_{-1}^{1} [g(x) - f(x)] dx = \int_{-1}^{1} [ 0 - (x^2 - 1) ] dx = \left( \frac{-x^3}{3} + x \right) \big|_{-1}^{1} = \frac{4}{3}\)

  \(A_3 = \int_{1}^{3} [f(x) - g(x)] dx = \int_{1}^{3} [(x^2 - 1) - 0] dx = \left( \frac{x^3}{3} - x \right) \big|_{1}^{3} = \frac{20}{3}\)

  Thus the area of the whole region between \(f\) and \(g\) on interval \([-2, 3]\) is

  \(A = A_1 + A_2 + A_3 = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}\)

  Caution : Remember, we should always first find all the subintervals on which \(f \ge g\) or \(f \le g\) before writing down the integral. If we derectly evaluate the integral of \(f-g\) on the whole interval \([-2, 3]\), then we get the following WRONG answer :

  \(\int_{-2}^{3} [ f(x) - g(x) ] dx\)

  \(= \int_{-2}^{3} [ (x^2 - 1) - 0 ] dx\)

  \(= \left( \frac{x^3}{3} - x \right) \big|_{-2}^{3}\)

  \(= \frac{20}{3}\).

• Example 3 : Endpoints of the Region Are NOT Given
  

  Find the area of the region bounded by the curves of \(f(x) = x^3 - 1\) and \(g(x) = x - 1\).

  In this example, the endpoints of the region bounded by \(f(x)\) and \(g(x)\) are NOT GIVEN and we need figure them out.

  Let \(f(x) = g(x)\). Then we have

  \(x^3 - 1 = x - 1\)

  \(\implies x^3 - x = 0\)

  \(\implies x(x^2 - 1) = 0\)

  \(\implies x(x+1)(x-1) = 0\)

  \(\implies x = 0, x = -1\) or \(x = 1\)

  so

  • the left endpoint of the region is \(a = \min \{ 0, -1, 1\} = -1\)
  • the right endpoint of the region is \(b = \max \{ 0, -1, 1\} = 1\)

  Therefore, the question is now rephrased as : find the area of the region between curves of \(f(x) = x^3 - 1\) and \(g(x) = x - 1\) on interval \([-1, 1]\).

  To calculate the area of whole region, we need first find all subintervals of \([-1, 1]\) on which \(f(x) \ge g(x)\) or \(f(x) \le g(x)\).

  • We have already found that \(f(x) = g(x)\) at \(x = -1\), \(x = 0\) and \(x = 1\)
  • Then we separate the whole interval \([-1, 1]\) by \(-1, 0\) and \(1\) into \((-1, 0)\) and \((0, 1)\)
  • Within \((-1, 0)\), we pick a test number \(d = -0.5\). Then \(f(-0.5) = (-0.5)^3 - 1 = -1.125\), \(g(-0.5) = -0.5 - 1 = -1.5\), \(f(-0.5) > g(-0.5)\) and thus \(f(x) > g(x)\) on \((-1, -0)\).
  • Within \((0, 1)\), we pick a test number \(d = 0.5\). Then \(f(0.5) = 0.5^3 - 1 = -0.875\), \(g(0.5) = 0.5 - 1 = -0.5\), \(f(0.5) < g(0.5)\) and thus \(f(x) < g(x)\) on \((0, 1)\).

  Then the areas of the regions on subintervals are

  \(A_1 = \int_{-1}^{0} [ f(x) - g(x) ] dx\) \(= \int_{-1}^{0} [ (x^3 - 1) - (x - 1) ] dx\) \(= \int_{-1}^{0} ( x^3 - x ) dx\) \(= \left( \frac{x^4}{4} - \frac{x^2}{2} \right) \big|_{-1}^{0}\) \(= \left( \frac{0^4}{4} - \frac{0^2}{2} \right) - \left( \frac{(-1)^4}{4} - \frac{(-1)^2}{2} \right)\) \(= \frac{1}{4}\)

  \(A_2 = \int_{0}^{1} [g(x) - f(x)] dx\) \(= \int_{0}^{1} [ (x - 1) - (x^3 - 1) ] dx\) \(= \int_{0}^{1} (x - x^3) dx\) \(= \left( \frac{x^2}{2} + \frac{x^4}{4} \right) \big|_{0}^{1}\) \(= \left( \frac{1^2}{2} + \frac{1^4}{4} \right) - \left( \frac{0^2}{2} + \frac{0^4}{4} \right)\) \(= \frac{3}{4}\)

  Thus the area of the whole region between \(f\) and \(g\) on interval \([-1, 1]\) is

  \(A = A_1 + A_2 = \frac{1}{4} + \frac{3}{4} = 1\)

4.7.1 Formula

4.7.2 Examples

• Example 1
  

  Find the volume of solid of revolution formed by rotating about \(x\)−axis the area bounded by \(f(x) = \sqrt[3]{2x+1} + 3\), \(x\)-axis, \(x = 1\) and \(x = 4\).

  We know \([f(x)]^2 = (\sqrt[3]{2x+1} + 3)^2 = [(2x+1)^{\frac{1}{3}} + 3]^2\)

  \(= [(2x+1)^{\frac{1}{3}}]^2 + 3^2 + 2 \cdot 3 \cdot (2x+1)^{\frac{1}{3}}\)

  \(= (2x+1)^{\frac{2}{3}} + 9 + 6(2x+1)^{\frac{1}{3}}\)

  Therefore, by the formula, the volume is

  \(V = \int_{1}^{4} \pi [f(x)]^2 dx\)

  \(= \pi \int_{1}^{4} \left[ (2x+1)^{\frac{2}{3}} + 9 + 6(2x+1)^{\frac{1}{3}} \right] dx\)

  \(= \pi \left[ \int_{1}^{4} (2x+1)^{\frac{2}{3}} dx + \int_{1}^{4} 9 dx + \int_{1}^{4} 6(2x+1)^{\frac{1}{3}} dx \right]\)

  \(= \pi \left[ \int_{1}^{4} (2x+1)^{\frac{2}{3}} dx + 9 \int_{1}^{4} 1 dx + 6 \int_{1}^{4} (2x+1)^{\frac{1}{3}} dx \right]\)

  \(= \pi (V_1 + 9V_2 + 6V_3)\)

  where

  \(V_1 = \int_{1}^{4} (2x+1)^{\frac{2}{3}} dx\)

  \(V_2 = \int_{1}^{4} 1 dx\)

  \(V_3 = \int_{1}^{4} (2x+1)^{\frac{1}{3}} dx\)

  • For \(V_1\):

    The integrand is a power whose base is another function, so we may use integration by substitution, by letting \(u\) be the base.

    Let \(u = 2x+1\). Then \(u'=(2x+1)'=2\) and \(du = u' dx = 2 dx\). Thus the indefinite integral is

    \(\int (2x+1)^{\frac{2}{3}} dx\) \(= \int (2x+1)^{\frac{2}{3}} \cdot \frac{2}{2} dx\) \(= \frac{1}{2} \int (2x+1)^{\frac{2}{3}} \cdot 2 dx\) \(= \frac{1}{2} \int u^{\frac{2}{3}} du\) \(= \frac{1}{2} \cdot \frac{u^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} + C\) \(= \frac{1}{2} \cdot \frac{3u^{\frac{5}{3}}}{5} + C\) \(= \frac{3u^{\frac{5}{3}}}{10} + C\) \(= \frac{3(2x+1)^{\frac{5}{3}}}{10} + C\)

    By FTC, we have

    \(V_1 = \int (2x+1)^{\frac{2}{3}} dx\) \(= \frac{3(2x+1)^{\frac{5}{3}}}{10} \big|_{1}^{4}\) \(= \frac{3(2\cdot 4+1)^{\frac{5}{3}}}{10} - \frac{3(2\cdot 1+1)^{\frac{5}{3}}}{10}\) \(= \frac{3\cdot 9^{\frac{5}{3}}}{10} - \frac{3\cdot 3^{\frac{5}{3}}}{10}\) \(= 9.81\)

  • For \(V_2\):

    \(\int 1 dx = x + C\)

    By FTC, we have

    \(V_2 = \int_{1}^{4} 1 dx\) \(= x \big|_{1}^{4}\) \(= 4 - 1 = 3\)

  • For \(V_3\):

    Similar to \(V_1\), the integrand is a power whose base is another function, so we may use integration by substitution, by letting \(u\) be the base.

    Let \(u = 2x+1\). Then \(u'=(2x+1)'=2\) and \(du = u' dx = 2 dx\). Thus the indefinite integral is

    \(\int (2x+1)^{\frac{1}{3}} dx\) \(= \int (2x+1)^{\frac{1}{3}} \cdot \frac{2}{2} dx\) \(= \frac{1}{2} \int (2x+1)^{\frac{1}{3}} \cdot 2 dx\) \(= \frac{1}{2} \int u^{\frac{1}{3}} du\) \(= \frac{1}{2} \cdot \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} + C\) \(= \frac{1}{2} \cdot \frac{3u^{\frac{4}{3}}}{4} + C\) \(= \frac{3u^{\frac{4}{3}}}{8} + C\) \(= \frac{3(2x+1)^{\frac{4}{3}}}{8} + C\)

    By FTC, we have

    \(V_3 = \int (2x+1)^{\frac{1}{3}} dx\) \(= \frac{3(2x+1)^{\frac{4}{3}}}{8} \big|_{1}^{4}\) \(= \frac{3(2\cdot 4+1)^{\frac{4}{3}}}{8} - \frac{3(2\cdot 1+1)^{\frac{4}{3}}}{8}\) \(= \frac{3\cdot 9 ^{\frac{4}{3}}}{8} - \frac{3\cdot 3^{\frac{4}{3}}}{8}\) \(= 5.40\)

  Putting \(V_1\), \(V_2\) and \(V_3\) back into the formula, then we get

  \(V = \int_{1}^{4} \pi [f(x)]^2 dx\) \(= \pi (V_1 + 9V_2 + 6V_3)\) \(= \pi (9.81 + 9\cdot 3 + 6\cdot 5.40)\) \(= 217.43\)

• Example 2
  

  Find the volume of solid of revolution formed by rotating about \(x\)−axis the area bounded by \(f(x) = e^{-2x+4} - 1\), \(x\)-axis, \(x = 1\) and \(x = 3\).

  We know \([f(x)]^2 = (e^{-2x+4} - 1)^2 = [e^{-2x+4} - 1]^2\)

  \(= (e^{-2x+4})^2 + 1^2 - 2 \cdot 1 \cdot e^{-2x+4}\)

  \(= e^{-4x+8} + 1 - 2 e^{-2x+4}\)

  Therefore, by the formula, the volume is

  \(V = \int_{1}^{3} \pi [f(x)]^2 dx\)

  \(= \pi \int_{1}^{3} \left[ e^{-4x+8} + 1 - 2 e^{-2x+4} \right] dx\)

  \(= \pi \left[ \int_{1}^{3} e^{-4x+8} dx + \int_{1}^{3} 1 dx - 2 \int_{1}^{3} e^{-2x+4} dx \right]\)

  \(= \pi (V_1 + V_2 - 2V_3)\)

  where

  \(V_1 = \int_{1}^{3} e^{-4x+8} dx\)

  \(V_2 = \int_{1}^{3} 1 dx\)

  \(V_3 = \int_{1}^{3} e^{-2x+4} dx\)

  • For \(V_1\):

    By Exponential Rule for integration, we have \(\int e^{-4x+8} dx\) \(= \frac{e^{-4x+8}}{-4} + C\)

    By FTC, we have

    \(V_1 = \int_{1}^{3} e^{-4x+8} dx\) \(= \frac{e^{-4x+8}}{-4} \big|_{1}^{3}\) \(= \frac{e^{-4\cdot 3 +8}}{-4} - \frac{e^{-4\cdot 1+8}}{-4}\) \(= \frac{e^{-4}}{-4} - \frac{e^{4}}{-4}\) \(= 13.65\)

  • For \(V_2\):

    \(\int 1 dx = x + C\)

    By FTC, we have

    \(V_2 = \int_{1}^{3} 1 dx\) \(= x \big|_{1}^{3}\) \(= 3 - 1 = 2\)

  • For \(V_3\):

    By Exponential Rule for integration, we have \(\int e^{-2x+4} dx\) \(= \frac{e^{-2x+4}}{-2} + C\)

    By FTC, we have

    \(V_3 = \int_{1}^{3} e^{-2x+4} dx\) \(= \frac{e^{-2x+4}}{-2} \big|_{1}^{3}\) \(= \frac{e^{-2\cdot 3+4}}{-2} - \frac{e^{-2\cdot 1+4}}{-2}\) \(= \frac{e^{-2}}{-2} - \frac{e^{2}}{-2}\) \(= 3.63\)

  Putting \(V_1\), \(V_2\) and \(V_3\) back into the formula, then we get

  \(V = \int_{1}^{4} \pi [f(x)]^2 dx\) \(= \pi (V_1 + V_2 - 2V_3)\) \(= \pi (13.65 + 2 - 2 \cdot 3.63)\) \(= 26.36\)

4.8.1 Formula

4.8.2 Example

5.1.1 Definition

5.1.2 Example

5.2.1 Informal Definition

5.2.2 Formal Definition

5.2.3 Remark

5.2.4 Example

5.3.1 Definitions

5.3.2 Remark

5.3.3 Example

5.4.1 Definition

5.4.2 Critical Point

• Definition
  

  For a function \(f(x,y)\), the points \((a,b)\) such that \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\) are called critical points.

• Remark
  

  AND condition: For a critical point \((a,b)\), \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\) should hold at the same time.

• How to find all critical points?
  
  1. Find all values of \(x\) such that \(f_x(x,y) = 0\) or \(f_y(x,y) = 0\), say \(x_1, x_2, \dots, x_m\)
  2. For \(x_1\)
     • put \(x=x_1\) into the equations \(f_x(x,y) = 0\) and \(f_y(x,y) = 0\), to get \(f_x(x_1,y) = 0\) and \(f_y(x_1,y) = 0\)
     • solve for all values of \(y\) such that \(f_x(x_1,y) = 0\) and \(f_y(x_1,y) = 0\), say \(y_{1},y_{2},\dots,y_{m}\)
     • then all \((x_1, y_{1}), (x_1, y_{2}), \dots, (x_1, y_{m})\) are critical points
  3. Repeat Step 2 for each \(x_2, \dots, x_n\), to get all the corresponding critical points.

  See Example 4 for illustration.

5.4.3 Saddle Point

5.4.4 Test for Relative Extrema

5.4.5 Example

• Example 1
  

  Find all relative minima, relative maxima or saddle points of \(f(x, y) = 8x^2 + 4xy + y^2 + 48x - 24y - 7\).

  First, find all partial derivatives and second-order partial derivatives.

  \(f_x(x,y) = 16x + 4y + 48\)

  \(f_y(x,y) = 4x + 2y -24\)

  \(f_{xx}(x,y) = 16\)

  \(f_{yy}(x,y) = 2\)

  \(f_{xy}(x,y) = f_{yx}(x,y) = 4\)

  Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

  \(16x + 4y + 48 = 0\),

  \(4x + 2y -24 = 0\).

  which means \(x = -12\) and \(y = 36\). Hence \((-12, 36)\) is the only critical point of function \(f\).

  Third, calculate the discriminant of \(f\) at the critical point \((-12, 36)\), ie,

  \(D = f_{xx}(-12, 36) \cdot f_{yy}(-12, 36) - [f_{xy}(-12, 36)]^2 = 16 \cdot 2 - 4^2 = 16\)

  Finally, since \(D > 0\) and \(f_{xx} > 0\), we know \((-12, 36)\) is a relative minimum of \(f\).

• Example 2
  

  Find all relative minima, relative maxima or saddle points of \(f(x, y) = -8x^2 + 8xy + y^2 + 48x - 24y - 7\).

  First, find all partial derivatives and second-order partial derivatives.

  \(f_x(x,y) = -16x + 8y + 48\)

  \(f_y(x,y) = 8x + 2y -24\)

  \(f_{xx}(x,y) = -16\)

  \(f_{yy}(x,y) = 2\)

  \(f_{xy}(x,y) = f_{yx}(x,y) = 8\)

  Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

  \(-16x + 8y + 48 = 0\),

  \(8x + 2y -24 = 0\).

  which means \(x = 3\) and \(y = 0\). Hence \((3, 0)\) is the only critical point of function \(f\).

  Third, calculate the discriminant of \(f\) at the critical point \((3, 0)\), ie,

  \(D = f_{xx}(3, 0) \cdot f_{yy}(3, 0) - [f_{xy}(3, 0)]^2 = -16 \cdot 2 - 8^2 = -96\)

  Finally, since \(D < 0\), we know \((3, 0)\) is a saddle point of \(f\).

• Example 3
  

  Find all relative minima, relative maxima or saddle points of \(f(x, y) = x^3 + 8y^3 - 48x - 24y - 7\).

  First, find all partial derivatives and second-order partial derivatives.

  \(f_x(x,y) = 3x^2 - 48 = 3(x^2 - 16) = 3(x + 4)(x - 4)\)

  \(f_y(x,y) = 24y^2 -24 = 24(y^2 - 1) = 24(y + 1)(y - 1)\)

  \(f_{xx}(x,y) = 6x\)

  \(f_{yy}(x,y) = 48y\)

  \(f_{xy}(x,y) = f_{yx}(x,y) = 0\)

  Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

  \(3(x + 4)(x - 4) = 0\),

  \(24(y + 1)(y - 1) = 0\).

  We see that \(-4\) and \(4\) are the only two values of \(x\) such that \(f_x = 0\) or \(f_y = 0\).

  • When \(x = -4\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. To have \(f_y = 0\), we need \(y = -1\) or \(y = 1\). Therefore, \((x,y) = (-4, -1)\) and \((x,y) = (-4, 1)\) are two critical points.
  • When \(x = 4\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. To have \(f_y = 0\), we need \(y = -1\) or \(y = 1\). Therefore, \((x,y) = (4, -1)\) and \((x,y) = (4, 1)\) are two critical points.

  Hence, there are 4 critical points in total : \((x,y) = (-4, -1), (-4, 1), (4, -1), (4, 1)\).

  Third, calculate the discriminant of \(f\) at each critical point:

  • at \((x,y) = (-4,-1)\), \(D = f_{xx}(-4, -1) \cdot f_{yy}(-4, -1) - [f_{xy}(-4, -1)]^2 = 1152 > 0\). Since \(f_{xx}(-4, -1) < 0\), \((x,y) = (-4, -1)\) yields a relative maximum
  • at \((x,y) = (-4,1)\), \(D = f_{xx}(-4, 1) \cdot f_{yy}(-4, 1) - [f_{xy}(-4, 1)]^2 = -1152 < 0\), so \((x,y) = (-4, 1)\) yields a saddle point
  • at \((x,y) = (4,-1)\), \(D = f_{xx}(4, -1) \cdot f_{yy}(4, -1) - [f_{xy}(4, -1)]^2 = -1152 < 0\), so \((x,y) = (4, -1)\) yields a saddle point
  • at \((x,y) = (4,1)\), \(D = f_{xx}(4, 1) \cdot f_{yy}(4, 1) - [f_{xy}(4, 1)]^2 = 1152 > 0\). Since \(f_{xx}(4, 1) > 0\), \((x,y) = (4, 1)\) yields a relative minimum.
• Example 4
  

  Find all relative minima, relative maxima or saddle points of \(f(x, y) = (x^2 - 1)(y^2 - 9)\).

  First, find all partial derivatives and second-order partial derivatives.

  \(f_x(x,y) = (y^2 - 9) \cdot 2x = 2x(y+3)(y-3)\)

  \(f_y(x,y) = (x^2 - 1) \cdot 2y = 2(x+1)(x-1)y\)

  \(f_{xx}(x,y) = 2(y^2 - 9)\)

  \(f_{yy}(x,y) = 2(x^2 - 1)\)

  \(f_{xy}(x,y) = f_{yx}(x,y) = 4xy\)

  Second, find all critical points by letting \(f_x = 0\) and \(f_y = 0\), ie,

  \(2x(y+3)(y-3) = 0\),

  \(2(x+1)(x-1)y = 0\).

  We see that \(0, -1\) and \(1\) are the values of \(x\) such that \(f_x = 0\) or \(f_y = 0\).

  • When \(x = 0\), we have \(f_x(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_y = 0\) hold. Putting \(x=0\) into the second equation, we get \(2(x+1)(x-1)y = 2\cdot(0+1)\cdot(0-1)\cdot y = -2y = 0\), which means \(y = 0\). Therefore \((x,y) = (0,0)\) is a critical point.
  • When \(x = -1\), we have \(f_y(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_x = 0\) hold. Putting \(x=-1\) into the first equation, we get \(2\cdot(-1)\cdot(y+3)(y-3) = -2(y+3)(y-3) = 0\), which means \(y = -3\) or \(y = 3\). Therefore \((x,y) = (-1, -3)\) and \((x,y) = (-1, 3)\) are two critical points.
  • When \(x = 1\), we have \(f_y(x,y) = 0\). A critical number should make both \(f_x = 0\) and \(f_y = 0\) hold at the same time. Hence let's see how to make \(f_x = 0\) hold. Putting \(x=1\) into the first equation, we get \(2\cdot 1\cdot(y+3)(y-3) = 2(y+3)(y-3) = 0\), which means \(y = -3\) or \(y = 3\). Therefore \((x,y) = (1, -3)\) and \((x,y) = (1, 3)\) are two critical points.

  Hence, there are 5 critical points in total : \((x,y) = (0,0), (-1, -3), (-1, 3), (1, -3), (1, 3)\).

  Third, calculate the discriminant of \(f\) at each critical point:

  • at \((x,y) = (0, 0)\), \(D = f_{xx}(0, 0) \cdot f_{yy}(0, 0) - [f_{xy}(0, 0)]^2 = 36 > 0\). Since \(f_{xx}(0, 0) = -18 < 0\), \((x,y) = (0, 0)\) yields a relative maximum
  • at \((x,y) = (-1,-3)\), \(D = f_{xx}(-1, -3) \cdot f_{yy}(-1, -3) - [f_{xy}(-1, -3)]^2 = -144 < 0\), so \((x,y) = (-1, -3)\) yields a saddle point
  • at \((x,y) = (-1, 3)\), \(D = f_{xx}(-1, 3) \cdot f_{yy}(-1, 3) - [f_{xy}(-1, 3)]^2 = -144 < 0\), so \((x,y) = (-1, 3)\) yields a saddle point
  • at \((x,y) = (1,-3)\), \(D = f_{xx}(1, -3) \cdot f_{yy}(1, -3) - [f_{xy}(1, -3)]^2 = -144 < 0\), so \((x,y) = (1, -3)\) yields a saddle point
  • at \((x,y) = (1,3)\), \(D = f_{xx}(1, 3) \cdot f_{yy}(1, 3) - [f_{xy}(1, 3)]^2 = -144 < 0\), so \((x,y) = (1, 3)\) yields a saddle point

5.5.1 Introduction

5.5.2 The Lagrange Multiplier Method

• Remark
  

  This method just helps us find relative extrema, not tells us whether the relative extremum is a relative minimum or a relative maximum.

5.5.3 Procedure of Applying Lagrange Multiplier

5.5.4 Solving the System of Equations by Substitution

5.5.5 Examples

• Example 1
  

  Find relative extrema of \(f(x,y) = xy\), subjuect to \(x + y = 10\).

  1. The constraint can be rewritten as \(g(x,y) = x + y - 10 = 0\).

  2. The Lagrange function is

     \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \( = xy - \lambda (x + y - 10)\) \( = xy - \lambda x - \lambda y + 10\lambda\).

  3. Find partial derivatives

     \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = y - \lambda\)

     \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x - \lambda\)

     \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - y + 10\)

  4. Form the system of equations

     System 1:

     \[ F_x(x,y,\lambda) = y - \lambda = 0 \ \ (1) \]

     \[ F_y(x,y,\lambda) = x - \lambda = 0 \ \ (2) \]

     \[ F_{\lambda}(x,y,\lambda) = -x - y + 10 = 0 \ \ (3) \]

  5. Let's solve the System 1 in Step 4 by substitution.

     • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = y\ \ (1')\).

     • Replace \(\lambda\) with \(y\) in the other two equations, and we get System 2.

       System 2:

       \[ x - y = 0 \ \ (2') \]

       \[ -x - y + 10 = 0 \ \ (3) \]

     Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

     • Represent \(y\) as a function of \(x\) : from \((2')\) we get \(y = x\ \ (2'')\).
     • Replace \(y\) with \(x\) in equation \((3)\), and we get \(- x - x + 10 = 0 \implies x = 5\)
     • By \((2'')\), we have \(y = x = 5\)

     Therefore, \((x,y) = (5,5)\) yields a relative extremum \(f(5,5) = 5 \cdot 5 =25\)

     Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

• Example 2
  

  Find relative extrema of \(f(x,y) = x^2 + 2y^2\), subjuect to \(3x + 4y = 5\).

  1. The constraint can be rewritten as \(g(x,y) = 3x + 4y - 5 = 0\).

  2. The Lagrange function is

     \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^2 + 2y^2 - \lambda (3x + 4y - 5)\) \(= x^2 + 2y^2 - 3 \lambda x - 4 \lambda y + 5 \lambda\).

  3. Find partial derivatives

     \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = 2x - 3\lambda\)

     \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = 4y - 4\lambda\)

     \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -3x - 4y + 5\)

  4. Form the system of equations

     System 1:

     \[ F_x(x,y,\lambda) = 2x - 3\lambda = 0 \ \ (1) \]

     \[ F_y(x,y,\lambda) = 4y - 4\lambda = 0 \ \ (2) \]

     \[ F_{\lambda}(x,y,\lambda) = -3x - 4y + 5 = 0 \ \ (3) \]

  5. Let's solve the System 1 in Step 4 by substitution.

     • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = \frac{2x}{3} \ \ (1')\).

     • Replace \(\lambda\) with \(\frac{2x}{3}\) in the other two equations, and we get System 2.

       System 2:

       \[ 4y - 4 \cdot \frac{2x}{3} = 4y - \frac{8x}{3} = 0 \ \ (2') \]

       \[ -3x - 4y + 5 = 0 \ \ (3) \]

     Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

     • Rrepresent \(y\) as a function of \(x\) : from \((2')\) we get \(y = \frac{2x}{3} \ \ (2'')\).
     • Replace \(y\) with \(\frac{2x}{3}\) in equation \((3)\), and we get \(- 3x - 4 \cdot \frac{2x}{3} + 5 = 0 \implies x = \frac{15}{17}\)
     • By \((2'')\), we have \(y = \frac{2}{3} \cdot \frac{15}{17} = \frac{10}{17}\)

     Therefore, \((x,y) = \left( \frac{15}{17},\frac{10}{17} \right)\) yields a relative extremum \(f\left( \frac{15}{17},\frac{10}{17} \right) = \frac{425}{289}\).

     Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

• Example 3
  

  Find relative extrema of \(f(x,y) = x^2 y\), subjuect to \(x + 2y = 3\).

  1. The constraint can be rewritten as \(g(x,y) = x + 2y - 3 = 0\).

  2. The Lagrange function is

     \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^2 y - \lambda (x + 2y - 3)\) \(= x^2 y - \lambda x - 2 \lambda y + 3 \lambda\).

  3. Find partial derivatives

     \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = 2xy - \lambda\)

     \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x^2 - 2\lambda\)

     \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - 2y + 3\)

  4. Form the system of equations

     System 1:

     \[ F_x(x,y,\lambda) = 2xy - \lambda = 0 \ \ (1) \]

     \[ F_y(x,y,\lambda) = x^2 - 2\lambda = 0 \ \ (2) \]

     \[ F_{\lambda}(x,y,\lambda) = -x - 2y + 3 = 0 \ \ (3) \]

  5. Let's solve the System 1 in Step 4 by substitution.

     • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = 2xy \ \ (1')\).

     • Replace \(\lambda\) with \(2xy\) in the other two equations, and we get System 2.

       System 2:

       \[ x^2 - 2 \cdot 2xy = x^2 - 4xy = 0 \ \ (2') \]

       \[ -x - 2y + 3 = 0 \ \ (3) \]

     Now we just need solve the System 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

     • Represent \(x\) as a function of \(y\) : from \((3)\) we get \(x = 3-2y \ \ (3')\).

     • Replace \(x\) with \(3-2y\) in equation \((3')\), and we get

       \((3-2y)^2 - 4(3-2y)y = 0\)

       \(\implies 3^2 + (2y)^2 - 2\cdot 3\cdot 2y - (12 - 8y)y = 0\)

       \(\implies 9 + 4y^2 - 12y - 12y + 8y^2 = 0\)

       \(\implies 12y^2 - 24y + 9 = 0\)

       \(\implies 4y^2 - 8y + 3 = 0\)

       \(\implies (2y-1)(2y-3) = 0\)

       \(y = \frac{1}{2}\) or \(y = \frac{3}{2}\)

     • By \((3')\), we have
       • when \(y = \frac{1}{2}\), \(x = 3 - 2 \cdot \frac{1}{2} = 2\)
       • when \(y = \frac{3}{2}\), \(x = 3 - 2 \cdot \frac{3}{2} = 0\)

     Therefore, \((x,y) = \left( 2,\frac{1}{2} \right)\) and \((x,y) = \left( 0,\frac{3}{2} \right)\) yield two relative extrema \(f\left( 2,\frac{1}{2} \right) = 2^2 \cdot \frac{1}{2} = 2\) and \(f\left( 0,\frac{3}{2} \right) = 0^2 \cdot \frac{3}{2} = 0\) respectively.

     Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

• Example 4
  

  Find relative extrema of \(f(x,y) = x^{\frac{1}{4}} y^{\frac{3}{4}}\), subjuect to \(x + 2y = 1\).

  1. The constraint can be rewritten as \(g(x,y) = x + 2y - 1 = 0\).

  2. The Lagrange function is

     \(F(x,y,\lambda) = f(x,y) - \lambda \cdot g(x,y)\) \(= x^{\frac{1}{4}} y^{\frac{3}{4}} - \lambda (x + 2y - 1)\) \(= x^{\frac{1}{4}} y^{\frac{3}{4}} - \lambda x - 2 \lambda y + \lambda\).

  3. Find partial derivatives

     \(F_x(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial x} = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} - \lambda\)

     \(F_y(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial y} = x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2\lambda\)

     \(F_{\lambda}(x,y,\lambda) = \frac{\partial F(x,y,\lambda)}{\partial \lambda} = -x - 2y + 1\)

  4. Form the system of equations

     System 1:

     \[ F_x(x,y,\lambda) = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} - \lambda = 0 \ \ (1) \]

     \[ F_y(x,y,\lambda) = x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2\lambda = 0 \ \ (2) \]

     \[ F_{\lambda}(x,y,\lambda) = -x - 2y + 1 = 0 \ \ (3) \]

  5. Let's solve the System 1 in Step 4 by substitution.

     • Represent \(\lambda\) as a function of \(x\) and \(y\) : from \((1)\) we get \(\lambda = \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \ \ (1')\).

     • Replace \(\lambda\) with \(\frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}}\) in the other two equations, and we get System 2.

       System 2:

       \[ x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} = 0 \ \ (2') \]

       \[ -x - 2y + 1 = 0 \ \ (3) \]

     Now we just need solve the system 2 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.

     • Before applying the substitution, let's simplify \((2')\) in some way. We see that \((2')\) has fraction powers and even negative fraction powers. Thus if we directly apply the substitution then those fraction powers still exist. We don't like fraction powers or negative powers, so let's eliminate them.
       • we see that the negative power of \(x\) in \((2')\) is \(\frac{-3}{4}\) and the negative power of \(y\) in \((2')\) is \(\frac{-1}{4}\).

       • so we multiply both sides of \((2')\) by \(x^{\frac{3}{4}} y^{\frac{1}{4}}\) in order to offset those negative powers

         \(x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} = 0 \ \ (2')\)

         \(\implies x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \right) = x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot 0\)

         \(\implies x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( x^{\frac{1}{4}} \cdot \frac{3}{4} \cdot y^{\frac{-1}{4}} \right) - x^{\frac{3}{4}} y^{\frac{1}{4}} \cdot \left( 2 \cdot \frac{1}{4} \cdot x^{\frac{-3}{4}} y^{\frac{3}{4}} \right) = 0\)

         \(\implies \frac{3}{4} \cdot x^{\frac{3}{4}} x^{\frac{1}{4}} y^{\frac{1}{4}} \cdot y^{\frac{-1}{4}} - 2 \cdot \frac{1}{4} \cdot x^{\frac{3}{4}} \cdot x^{\frac{-3}{4}} \cdot y^{\frac{1}{4}} \cdot y^{\frac{3}{4}} = 0\)

         \(\implies \frac{3}{4} \cdot x - 2 \cdot \frac{1}{4} \cdot y = 0\)

         \(\implies y = \frac{3x}{2} \ \ (2'')\)

         Note here we don't have any fraction power or negative power any more.

       • Now System 2 becomes

         new System 2:

         \[ y = \frac{3x}{2} \ \ (2'') \]

         \[ -x - 2y + 1 = 0 \ \ (3) \]

     • Now let's solve the new System 2 by substitution.
       • \(y\) is already represent as \(\frac{3x}{2}\) in \((2'')\)

       • Replace \(y\) with \(\frac{3x}{2}\) in equation \((3)\), and we get

         \(-x - 2 \cdot \frac{3x}{2} + 1 = 0\)

         \(\implies -x - 3x + 1 = 0\)

         \(\implies x = \frac{1}{4}\)

       • By \((2'')\), we have \(y = \frac{3}{2} \cdot \frac{1}{4} = \frac{3}{8}\)

     Therefore, \((x,y) = \left( \frac{1}{4}, \frac{3}{8} \right)\) yields a relative extremum \(f\left( \frac{1}{4}, \frac{3}{8} \right) = \left( \frac{1}{4} \right) ^ {\frac{1}{4}} \left( \frac{3}{8} \right) ^ {\frac{3}{4}}\).

     Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

• Example 5
  

  Find relative extrema of \(f(x,y,z) = xyz\), subjuect to \(2xy + 2xz + yz = 12\).

  1. The constraint can be rewritten as \(g(x,y,z) = 2xy + 2xz + yz - 12 = 0\).

  2. The Lagrange function is

     \(F(x,y,z,\lambda) = f(x,y,z) - \lambda \cdot g(x,y,z)\) \(= xyz - \lambda (2xy + 2xz + yz - 12)\) \(= xyz - 2\lambda xy - 2\lambda xz - \lambda yz + 12\lambda\)

  3. Find partial derivatives

     \(F_x(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial x} = yz - 2\lambda y - 2\lambda z\)

     \(F_y(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial y} = xz - 2\lambda x - \lambda z\)

     \(F_z(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial z} = xy - 2\lambda x - \lambda y\)

     \(F_{\lambda}(x,y,z,\lambda) = \frac{\partial F(x,y,z,\lambda)}{\partial \lambda} = - 2xy - 2xz - yz + 12\)

  4. Form the system of equations

     System 1:

     \[ F_x(x,y,z,\lambda) = yz - 2\lambda y - 2\lambda z = 0 \ \ (1) \]

     \[ F_y(x,y,z,\lambda) = xz - 2\lambda x - \lambda z = 0 \ \ (2) \]

     \[ F_z(x,y,z,\lambda) = xy - 2\lambda x - \lambda y = 0 \ \ (3) \]

     \[ F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4) \]

  5. NOTE : the hierarchical structure of the following solution is indicated by the bullet symbol \(\bullet\) and indentation.

     Let's solve the System 1 in Step 4 by substitution.

     • Represent \(\lambda\) as a function of \(x, y\) and \(z\) : from \((1)\) we get \(\lambda = \frac{yz}{2y + 2z} \ \ (1')\).

     • Replace \(\lambda\) with \(\frac{yz}{2y + 2z}\) in the other three equations, and we get

       \(F_y(x,y,z,\lambda) = xz - 2\cdot \frac{yz}{2y + 2z} \cdot x - \frac{yz}{2y + 2z} \cdot z\) \(= \frac{xz(2y+2z)}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{yz^2}{2y + 2z}\) \(= \frac{2xyz+2xz^2}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{yz^2}{2y + 2z}\) \(= \frac{2xyz+2xz^2 - 2xyz - yz^2}{2y + 2z}\) \(= \frac{(2x - y)z^2}{2y + 2z} = 0\)

       \(\implies (2x - y)z^2 = 0 \ \ (2')\)

       \(F_z(x,y,z,\lambda) = xy - 2\cdot \frac{yz}{2y + 2z} \cdot x - \frac{yz}{2y + 2z} \cdot y\) \(= \frac{xy(2y+2z)}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{y^2 z}{2y + 2z}\) \(= \frac{2xy^2+2xyz}{2y + 2z} - \frac{2xyz}{2y + 2z} - \frac{y^2 z}{2y + 2z}\) \(= \frac{2xy^2+2xyz - 2xyz - y^2 z}{2y + 2z}\) \(= \frac{(2x - z)y^2}{2y + 2z} = 0\)

       \(\implies (2x - z)y^2 = 0 \ \ (3')\)

       \(F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4)\)

       Putting \((2')\), \((3')\) and \((4)\) together, we have System 2 which contains only \(x, y\) and \(z\) as follows.

       System 2:

       \[ (2x - y)z^2 = 0 \ \ (2') \]

       \[ (2x - z)y^2 = 0 \ \ (3') \]

       \[ F_{\lambda}(x,y,z,\lambda) = - 2xy - 2xz - yz + 12 = 0 \ \ (4) \]

     Now we just need solve the system 2, which contains only \(x, y\) and \(z\). Again, we use substitution method to solve it.

     From \((2')\), we get

     \[ z = 0 \ \ (5) \textbf{ OR } 2x - y = 0 \implies y = 2x \ \ (6) \]

     Note the relationship between \((5)\) and \((6)\) is OR, not AND, so we don't put them simultaneously into equations \((3')\) and \((4)\) in System 2. Instead, at each time, we just use one of them. As a result, we need discuss it case by case.

     • Case 1 : When \(z = 0 \ \ (5)\) holds

       • Replace \(z\) with \(0\) in equations \((3')\) and \((4)\), and we get System 3.1.

         System 3.1:

         \[ 2xy^2 = 0 \ \ (3'') \]

         \[ -2xy + 12 = 0 \ \ (4') \]

       • Now we just need solve the system 3.1 which contains only \(x\) and \(y\). Again, we use substitution method to solve it.
         • From \((3'')\) we get \[x = 0 \ \ (7) \textbf{ OR } y^2 = 0 \implies y = 0 \ \ (8)\]
         • When \(x = 0 \ \ (7)\) holds, replace \(x\) with \(0\) in equation \((4')\), and we get \(12 = 0\), which never holds, so there is no solution of \(y\) corresponding to \(x=0\)
         • When \(y = 0 \ \ (8)\) holds, replace \(y\) with \(0\) in equation \((4')\), and we get \(12 = 0\), which never holds, so there is no solution of \(x\) corresponding to \(y=0\)
         • Therefore, there is not solution to System 3.1 under either \((7)\) or \((8)\).
       • Therefore, there is no solution to \((x,y,z)\) in Case 1.
     • Case 2 : When \(y = 2x \ \ (6)\) holds

       • Replace \(y\) with \(2x\) in equations \((3')\) and \((4)\), and we get System 3.2.

         System 3.2:

         \[ (2x - z) (2x)^2 = 4(2x - z)x^2 = 0 \ \ (3''') \]

         \[ -2x(2x) - 2xz - 2xz + 12 = -4x^2 - 4xz + 12 = 0 \ \ (4'') \]

       • Now we just need solve the system 3.2 which contains only \(x\) and \(z\). Again, we use substitution method to solve it.

         • From \((3''')\), we get

           \[ x^2 = 0 \implies x = 0 \ \ (9) \textbf{ OR } 2x - z = 0 \implies z = 2x \ \ (10) \]

           Note that the relationship between \((9)\) and \((10)\) is OR, not AND, so we don't put them simultaneously into equations \((4'')\) in System 3.2. Instead, at each time, we just use one of them. As a result, we need discuss it case by case.

         • Case 2.1 : When \(x = 0 \ \ (9)\) holds

           Replace \(x\) with \(0\) in equation \((4'')\), and we get \(12 = 0\), which never holds, so there is no solution of \(z\) corresponding to \(x = 0\).

           Therefore, there is no solution to \((x,y,z)\) in Case 2.1.

         • Case 2.2 : When \(z = 2x \ \ (10)\) holds

           • Replace \(z\) with \(2x\) in equation \((4'')\), and we get \(-4x^2 - 4x(2x) + 12 = 0\), so

             \[ x = 1 \ \ (11) \textbf{ OR } x = -1 \ \ (12) \]

           • When \(x = 1 \ \ (11)\) holds, by \((10)\) we have \(z = 2x = 2 \cdot 1 = 2\) and by \((6)\) we have \(y = 2x = 2 \cdot 1 = 2\). Therefore, \((x,y,z) = (1, 2, 2)\) gives a relative extremum.
           • When \(x = -1 \ \ (12)\) holds, by \((10)\) we have \(z = 2x = 2 \cdot (-1) = -2\) and by \((6)\) we have \(y = 2x = 2 \cdot (-1) = -2\). Therefore, \((x,y,z) = (-1, -2, -2)\) gives a relative extremum.
           • Therefore, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only two points giving relative extrema in Case 2.2.
       • Also, in whole Case 2, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only 2 points giving relative extrema as Case 2.1 has no solution.
     • Finally, for the whole problem, \((x,y,z) = (1,2,2)\) and \((x,y,z) = (-1,-2,-2)\) are the only 2 points giving relative extrema as Case 1 has no solution. The values of the two extrema are \(f(1,2,2) = 1\cdot 2\cdot 2 = 4\) and \(f(-1, -2, -2) = (-1) \cdot (-2) \cdot (-2) = -4\).

     Remark: You can use any method to solve these equations. Here I just show you how to apply the subtitution method. It may not be the simplest method, but it always works as far as this course is concerned. Using this "stupid" substitution method sometimes has an advantage: at least you will get the answer even if you are not good at manipulating complicated algebraic operations and tricks.

5.6.1 Total Differential for Two Variables

5.6.2 Approximations

• Remark
  

  This can be derived as follows. First, we have \(\Delta z \approx dz \ \ (1)\)

  \(\Delta z = f(x+dx,y+dy) - f(x,y) \ \ (2)\)

  \(dz = f_x(x,y) dx + f_y(x,y) dy \ \ (3)\)

  Then putting \((2)\) and \((3)\) into \((1)\) we get

  \(f(x+dx,y+dy) - f(x,y) \approx f_x(x,y) dx + f_y(x,y) dy\)

  \(\implies f(x+dx,y+dy) \approx f(x,y) + f_x(x,y) dx + f_y(x,y) dy\).

5.6.3 Example

• Example 1
  

  Find the total differential of function \(z = x^2 y\) and compute its values when \((x, y) = (2, 3)\) and \((dx, dy) = (\Delta x, \Delta y) = (0.01, 0.02)\).

  Find the partial derivatives

  \(f_x(x,y) = \frac{\partial f(x,y)}{\partial x} = \frac{\partial x^2 y}{\partial x} = 2xy\)

  \(f_x(x,y) = \frac{\partial f(x,y)}{\partial y} = \frac{\partial x^2 y}{\partial x} = x^2\)

  Then the total differential is \(dz = f_x(x,y) dx + f_y(x,y) dy = 2xy dx + x^2 dy\)

  When \((x,y) = (2,3)\) and \((dx, dy) = (0.01, 0.02)\), we have \(dz = f_x(x,y) dx + f_y(x,y) dy = 2xy dx + x^2 dy\) \(= 2 \cdot 2 \cdot 3 \cdot 0.01 + 2^2 \cdot 0.02\) \(= 0.2\)

• Example 2
  

  Approximate \(2.02^2 \cdot 2.99\).

  Let's identify the relevant quantities first:

  • the function is \(f(x,y) = x^2 y\)
  • we want to approximate \(f(2.02, 2.99)\)
  • \((2.02, 2.99)\) is close to \((2, 3)\) and \(f(2,3)\) is easy to compute, so we should approximate \(f(2.02, 2.99)\) by \(f(2,3)\)
  • \(dx = 2.02 - 2 = 0.02\) and \(dy = 2.99 - 3 = -0.01\)

  Then calculate

  • \(f_x(x,y) = \frac{\partial x^2 y}{\partial x} = 2xy\)
  • \(f_x(x,y) = \frac{\partial x^2 y}{\partial y} = x^2\)

  Therefore, by the formula \(f(x+dx, y+dy) \approx f(x,y) + f_x(x,y) dx + f_y(x,y) dy\), we have

  \(f(2.02, 2.99) = f(2+0.02, 3+(-0.01))\)

  \(\approx f(2,3) + f_x(2,3) \cdot 0.02 + f_y(2,3) \cdot (-0.01)\)

  \(= 2^2 \cdot 3 + 2 \cdot 2 \cdot 3 \cdot 0.02 + 2^2 \cdot (-0.01)\)

  \(= 12.2\)

5.7.1 Indefinite Integral

• Example
  

  Let \(f(x,y) = x^2 + xy + y^3\). Then

  \(\int f(x,y) dx = \int (x^2 + xy + y^3) dx = \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y)\) where \(C(y)\) is any function of \(y\).

  \(\int f(x,y) dy = \int (x^2 + xy + y^3) dy = x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x)\) where \(C(x)\) is any function of \(x\).

• Remark
  
  • \(\int f(x,y) dx \neq \int f(x,y) dy\). That is, the integral of \(f\) with respect to \(x\) is not identical to the integral with respect to \(y\). This is the general situation.
  • As we did for univariate functions, the indefinite integral, as the "most general antiderivative", is a class of antiderivatives rather than a single one. One antiderivative in this class can be represented as another one plus a constant.
  • Here the "constant" means a function that doesn't have the variable with respect to which the integral is defined. That is, it can be a function of the other variables. To find \(\int f(x,y) dx\), we add \(C(y)\) (an arbitrary function of \(y\)) to one specific antiderivative of function \(f(x,y)\) with respect to \(x\) (ie, \(\frac{x^3}{3} + \frac{yx^2}{2} + y^3 x\)), and the result is just the whole family of antiderivatives of \(f(x,y)\) with respect to \(x\). Check it by finding the partial derivative of \(\frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y)\) with respect to \(x\). The similar situation for \(\int f(x,y) dy\), the indefinite integral with respect to \(y\).
  • In \(\int f(x,y) dx\), if we use just a constant number \(C\) instead of \(C(y)\) (an arbitrary function of \(y\)), then the result is still a family of antiderivatives of \(f(x,y)\) with respect to \(x\). However, this family is not "the most general" one – not as general as the one using \(C(y)\) – so it is not the "indefinite integral".

5.7.2 Definite Integral

• Example
  

  Let \(f(x,y) = x^2 + xy + y^3\). Then

  \(\int_0^1 f(x,y) dx = \int_0^1 (x^2 + xy + y^3) dx\)

  \(= \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \big|_0^1\)

  \(= \left( \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \right) \big|_{x=1} - \left( \frac{x^3}{3} + \frac{yx^2}{2} + y^3 x + C(y) \right) \big|_{x=0}\)

  \(= \left( \frac{1^3}{3} + \frac{y \cdot 1^2}{2} + y^3 \cdot 1 + C(y) \right) - \left( \frac{0^3}{3} + \frac{y \cdot 0^2}{2} + y^3 \cdot 0 + C(y) \right)\)

  \(= \frac{1}{3} + \frac{y}{2} + y^3\)

  \(\int_0^1 f(x,y) dy = \int_0^1 (x^2 + xy + y^3) dy\)

  \(= x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \big|_0^1\)

  \(= \left( x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \right) \big|_{y=1} - \left( x^2 y + \frac{xy^2}{2} + \frac{y^4}{4} + C(x) \right) \big|_{y=0}\)

  \(= \left( x^2 \cdot 1 + \frac{x \cdot 1^2}{2} + \frac{1^4}{4} + C(x) \right) - \left( x^2 \cdot 0 + \frac{x \cdot 0^2}{2} + \frac{0^4}{4} + C(x) \right)\)

  \(= x^2 + \frac{x}{2} + \frac{1}{4}\)

• Remark
  
  • \(\int_0^1 f(x,y) dx \neq \int_0^1 f(x,y) dy\). This is the general situation, as the indefinite integrals \(\int f(x,y) dx \neq \int f(x,y) dy\).
  • \(\int_0^1 f(x,y) dx\) is a function of only \(y\), and \(\int_0^1 f(x,y) dy\) is a function of only \(x\). This is the general situation. When calculating the definite integral with respect to \(x\), we replace \(x\) in the antiderivative with the lower and upper integral limits, so \(x\) is eliminated and the result \(\int_0^1 x^2 + xy + y^3 dx = \frac{1}{3} + \frac{y}{2} + y^3\) contains only \(y\). It can be imagined that if we further find the definite integral of \(\frac{1}{3} + \frac{y}{2} + y^3\) with respect to \(y\) (ie, find \(\int_a^b \frac{1}{3} + \frac{y}{2} + y^3 dy\)), then \(y\) will be also eliminated and we will get a constant number. We will see this in next section "Double Integral".
  • When we evaluate \(\int_0^1 f(x,y) dx\), the definite integral of \(f(x,y)\) with respect to \(x\), \(C(y)\) doesn't play a role because it is eventually cancelled by \(-C(y)\). This is the general case : we don't really need \(C(y)\) for evaluating the definite integral with respect to \(x\), just as we don't need \(C\) for evaluating the univariate definite integral. The similar for the definite integral with respect to \(y\) and \(C(x)\).

5.8.1 Introduction: Volume of The Solid Under A Surface and Over A Region

5.8.2 Double Integral Over A Rectangle Region

• Formula
  

  For a non-negative \(f(x,y)\), if \(R = \{ (x,y) : a \le x \le b, c \le y \le d \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by either \(\iint_R f(x,y) dy dx\) or \(\iint_R f(x,y) dx dy\), equals either \(\int_a^b \left( \int_c^d f(x,y) dy \right) dx\) or \(\int_c^d \left( \int_a^b f(x,y) dx \right) dy\). That is,

  \[ V = \iint_R f(x,y) dy dx = \iint_R f(x,y) dx dy \] \[ = \int_a^b \left( \int_c^d f(x,y) dy \right) dx = \int_c^d \left( \int_a^b f(x,y) dx \right) dy \]

• Remark
  

  For rectangular \(R\), we always have \(\int_a^b \left( \int_c^d f(x,y) dy \right) dx = \int_c^d \left( \int_a^b f(x,y) dx \right) dy\), which means the order of the two layers of integrals can be exchanged.

  

  source of picture:
  https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/255doub/plot1.gif

• Example
  

  Find the volume of the solid under the graph of the function \(f(x,y) = 6x^2 y + e^{x+2y}\) and over the region \(R = \{ (x, y) | 3 \le x \le 5, 1 \le y \le 2 \}\).

  To calculate the volume, we need find either \(\int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx\) or \(\int_1^2 \left( \int_3^5 6x^2 y + e^{x+2y} dx \right) dy\).

  Let's find \(\int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx\) now. You can find \(\int_1^2 \left( \int_3^5 6x^2 y + e^{x+2y} dx \right) dy\) by yourself using similar steps.

  First, find the inner integral \(\int_1^2 6x^2 y + e^{x+2y} dy\).

  • Regarding \(x\) as a constant, we get the indefinite integral of \(6x^2 y + e^{x+2y}\) with respect to \(y\)

    \(\int 6x^2 y + e^{x+2y} dy\)

    \(= \int 6x^2 y dy + \int e^{x+2y} dy\)

    \(= 6x^2 \cdot \frac{y^2}{2} + \frac{e^{x+2y}}{2} + C(x)\)

    \(= 3 x^2 y^2 + \frac{e^{x+2y}}{2} + C(x)\)

    Hence, by FTC we get the definite integral \(\int_1^2 6x^2 y + e^{x+2y} dy\)

    \(= \left( 3 x^2 y^2 + \frac{e^{x+2y}}{2} \right) \big|_{y=1}^{y=2}\)

    \(= \left( 3 x^2 \cdot 2^2 + \frac{e^{x + 2 \cdot 2}}{2} \right) - \left( 3 x^2 \cdot 1^2 + \frac{e^{x+2 \cdot 1}}{2} \right)\)

    \(= \left( 12 x^2 + \frac{e^{x + 4}}{2} \right) - \left( 3 x^2 + \frac{e^{x+2}}{2} \right)\)

    \(= 12 x^2 + \frac{e^{x + 4}}{2} - 3 x^2 - \frac{e^{x + 2}}{2}\)

    \(= 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2}\)

  Second, find the outer integral \(\int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\).

  • The indefinite integral

    \(\int \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\)

    \(= \int 9 x^2 dx + \frac{1}{2} \int e^{x + 4} dx - \frac{1}{2} \int e^{x + 2} dx\)

    \(= 3 x^3 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} + C\)

  • Hence, by FTC we get the definite integral

    \(\int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx\)

    \(= \left( 3 x^3 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) \big|_{x=3}^{x=5}\)

    \(= \left( 3 \cdot 5^3 + \frac{e^{5 + 4}}{2} - \frac{e^{5 + 2}}{2} \right) - \left( 3 \cdot 3^3 + \frac{e^{3 + 4}}{2} - \frac{e^{3 + 2}}{2} \right)\)

    \(= 294 + \frac{e^9 - e^7 - e^7 + e^5}{2}\)

    \(= 294 + \frac{e^9 - 2e^7 + e^5}{2}\)

  Therefore, we have \(V = \iint_R f(x,y) dy dx = \int_3^5 \left( \int_1^2 6x^2 y + e^{x+2y} dy \right) dx = \int_3^5 \left( 9 x^2 + \frac{e^{x + 4}}{2} - \frac{e^{x + 2}}{2} \right) dx = 294 + \frac{e^9 - 2e^7 + e^5}{2}\)

5.8.3 Double Integral Over A Variable Region

• Formula
  
  • Case 1 : For a non-negative \(f(x,y)\), if the region \(R = \{ (x,y) : a \le x \le b, g_1(x) \le y \le g_2(x) \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by \(\iint_R f(x,y) dy dx\), equals \(\int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx\). That is \[ V = \iint_R f(x,y) dy dx = \int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx \]

  • Case 2 : For a non-negative \(f(x,y)\), if the region \(R = \{ (x,y) : h_1(y) \le x \le h_2(y), c \le y \le d \}\), then the volume of the solid under the graph of \(f\) and over the region \(R\), denoted by \(\iint_R f(x,y) dx dy\), equals \(\int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy\). That is

    \[ V = \iint_R f(x,y) dx dy = \int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy \]

• Remark
  
  • For non-rectangular \(R\), \(\int_a^b \left[ \int_{g_1(x)}^{g_2(x)} f(x,y) dy \right] dx \neq \int_{g_1(x)}^{g_2(x)} \left[ \int_a^b f(x,y) dx \right] dy\), and \(\int_c^d \left[ \int_{h_1(y)}^{h_2(y)} f(x,y) dx \right] dy \neq \int_{h_1(y)}^{h_2(y)} \left[ \int_c^d f(x,y) dx \right] dy\), which means the order of the two layers of integrals can NOT be exchanged.
  • When the integral limits of \(y\) are functions of \(x\), we should first calculate \(\int_{g_1(x)}^{g_2(x)} f(x,y) dy\), and then calculate the integral of the whole \(\int_{g_1(x)}^{g_2(x)} f(x,y) dy\) with respect to \(x\).
  • When the integral limits of \(x\) are functions of \(y\), we should first calculate \(\int_{h_1(y)}^{h_2(y)} f(x,y) dx\), and then calculate the integral of the whole \(\int_{h_1(y)}^{h_2(y)} f(x,y) dx\) with respect to \(y\).
  • For short, always let the inside integral be the one with variable integral limits.

  

  

  source of picture:
  https://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx

• Example
  

  Find the volume of the solid under the graph of the function \(f(x,y) = 6x^2 y + e^{x+2y}\) and over the region \(R = \{ (x, y) | y \le x \le 2y, 1 \le y \le 2 \}\).

  Note that here the integral limits of \(x\) are functions of \(y\) and the integral limits of \(y\) are constants, so to calculate the volume, we need first do the integration with respect to \(x\) and then do the integration with respect to \(y\). That is, we need find \(\int_1^2 \left( \int_{y}^{2y} 6x^2 y + e^{x+2y} dx \right) dy\).

  First, find the inner integral \(\int_{y}^{2y} 6x^2 y + e^{x+2y} dx\).

  • Regarding \(y\) as a constant, we get the indefinite integral of \(6x^2 y + e^{x+2y}\) with respect to \(x\)

    \(\int 6x^2 y + e^{x+2y} dx\)

    \(= \int 6x^2 y dx + \int e^{x+2y} dx\)

    \(= 6y \cdot \frac{x^3}{3} + e^{x+2y} + C(y)\)

    \(= 2 x^3 y + e^{x+2y} + C(y)\)

  • Hence, by FTC we get the definite integral

    \(\int_{y}^{2y} 6x^2 y + e^{x+2y} dx\)

    \(= ( 2 x^3 y + e^{x+2y} ) \big|_{x = y}^{x = 2y}\)

    \(= ( 2 \cdot (2y)^3 y + e^{2y+2y} ) - ( 2 y^3 y + e^{y+2y} )\)

    \(= ( 16 y^4 + e^{4y} ) - ( 2 y^4 + e^{3y} )\)

    \(= 16 y^4 + e^{4y} - 2 y^4 - e^{3y}\)

    \(= 14 y^4 + e^{4y} - e^{3y}\)

  Second, find the outer integral \(\int_1^2 \left( 14 y^4 + e^{4y} - e^{3y} \right) dy\)

  • The indefinite integral

    \(\int 14 y^4 + e^{4y} - e^{3y} dy\)

    \(= \int 14 y^4 dy + \int e^{4y} dy - \int e^{3y} dy\)

    \(= \frac{14y^5}{5} + \frac{e^{4y}}{4} - \frac{e^{3y}}{3} + C\)

  • Hence by FTC we get the definite integral

    \(\int_1^2 14 y^4 + e^{4y} - e^{3y} dx\)

    \(= \left( \frac{14y^5}{5} + \frac{e^{4y}}{4} - \frac{e^{3y}}{3} \right) \big|_{y=1}^{y=2}\)

    \(= \left( \frac{14 \cdot 2^5}{5} + \frac{e^{4 \cdot 2}}{4} - \frac{e^{3 \cdot 2}}{3} \right) - \left( \frac{14 \cdot 1^5}{5} + \frac{e^{4 \cdot 1}}{4} - \frac{e^{3 \cdot 1}}{3} \right)\)

    \(= \left( \frac{448}{5} + \frac{e^{8}}{4} - \frac{e^{6}}{3} \right) - \left( \frac{14}{5} + \frac{e^{4}}{4} - \frac{e^{3}}{3} \right)\)

    \(= \frac{448}{5} + \frac{e^{8}}{4} - \frac{e^{6}}{3} - \frac{14}{5} - \frac{e^{4}}{4} + \frac{e^{3}}{3}\)

    \(= \frac{434}{5} + \frac{e^{8} - e^{4}}{4} - \frac{e^{3} - e^{6}}{3}\)

  Therefore, we have \(V = \int_1^2 \left( \int_{y}^{2y} 6x^2 y + e^{x+2y} dx \right) dy = \int_1^2 \left( 14 y^4 + e^{4y} - e^{3y} \right) dy = \frac{434}{5} + \frac{e^{8} - e^{4}}{4} - \frac{e^{3} - e^{6}}{3}\)

6.1.1 Motivation

6.1.2 General Solution and Particular Solution

6.2.1 Formula

6.2.2 Example

6.3.1 Formula

6.3.2 Remark

• Why is it called separable?
  

  The differential equation is called "separable" because the right hand side \(\frac{p(x)}{q(y)}\) is a fraction between \(p(x)\) and \(q(y)\), where

  • \(p(x)\) is a function of ONLY \(x\)
  • \(q(y)\) is a function of ONLY \(y\)

  That means, the two variables \(x\) and \(y\) are separated into two individual functions.

• A Common Mistake
  

  Find \(\frac{dy}{dx}=\frac{1}{xy}\).

  Wrong Solution:

  let \(p(x) = 1\) and \(q(y)=xy\).

  Then the general solution is

  \(\int q(y) dy = \int p(x) dx\)

  \(\implies \int xy dy = \int 1 dx\)

  \(\implies x \int y dy = \int 1 dx\)

  \(\implies x \cdot \frac{y^2}{2} + C_1 = x + C_2\)

  \(\implies \frac{xy^2}{2} - x = C\)

  What is wrong with the following solution?

  • First, as mentioned above, \(p(x)\) should be a function of ONLY \(x\), and \(q(y)\) should be a function of ONLY \(y\). However, \(q(y)=xy\) in the above wrong solution was defined as a function of both \(x\) and \(y\). That means, \(x\) and \(y\) were not separated. The correct way is \(\frac{dy}{dx}\) \(= \frac{1}{xy}\) \(= \frac{1}{x} \cdot \frac{1}{y}\) \(= \frac{\frac{1}{x}}{y}\) \(= frac{p(x)}{q(y)}\), ie, let \(p=\frac{1}{x}\) and let \(q(y) = y\).

  • Second, remember that here \(\int xy dy \neq x \int y dy\)! Why? We did have \(\int xy dy = x \int y dy\) in the multivariate calculus chapter! Yes, in the multivariate calculus, we had \(\int xy dy = x \int y dy\) because we regard \(x\) as a constant when \(y\) is regarded as the variable. Why are we allowed to do so (regarding \(x\) as a constant)? The reason is, in a multivariate function \(z=f(x,y)\), \(x\) and \(y\) are independent of each other. That is, \(x\) doesn't change as \(y\) changes, so \(x\) is "constant".

    However, when saying "differential equation", we mean \(y = f(x)\) is a univariate function of \(x\), rather than any multivariate function \(z=f(x,y)\)! That is, \(x\) and \(y\) are not indpendent! Suppose \(y\) is actually independent of \(x\). Then \(y\) can be regarded as a constant when \(x\) is regarded as the variable, and thus always \(\frac{dy}{dx} = \frac{d(constant)}{dx} = 0\), which makes the diffrential equation meaningless!

    To summarize,

    • If we are talking about the multivariate function \(z = f(x,y)\), in which \(z\) is a function of both \(x\) and \(y\) and \(x\) and \(y\) are independent, then we can regard \(x\) as a constant when \(y\) is treated as the variable. Also, we can regard \(y\) as a constant when \(x\) is treated as the variable.
    • If we are talking about (ordinary) differential equations, then we mean \(y\) is a univariate function of \(x\), ie, \(y=f(x)\), and thus \(x\) and \(y\) are not independent of each other. In this case, we can NOT regard \(x\) as a constant when \(y\) is treated as the variable. Also, we can NOT regard \(y\) as a constant when \(x\) is treated as the variable.

    Let's look at the following example to get an illustration.

    • Case 1 : multivariate \(z = f(x,y) = xy\). Then \(x\) and \(y\) are independent, so \(y\) can be regarded as a constant when \(x\) is regarded as the variable. In this case, we have \(\int xy dx = y \int x dx = y \left( \frac{x^2}{2} + C \right) = \frac{x^2 y}{2} + Cy\)
    • Case 2 : univariate \(y = f(x) = x^3\). Then \(y\) depends on \(x\), and can NOT be regarded as a constant. In this case, we have \(\int xy dx = \int x \cdot x^3 dx = \int x^4 dx = \frac{x^5}{5} + C = \frac{x^2 \cdot x^3}{5} + C = \frac{x^2 y}{5} + C\).
    • Look, for Case 1 and Case 2, the answers to \(\int xy dx\) are not the same, because they are based on different assumptions!

6.3.3 Example 1

6.3.4 Example 2

6.3.5 Example 3

6.4.1 Definition

6.4.2 Integrating Factor

6.4.3 Solving The Equation

6.4.4 Examples

• Example 1
  

  Find the particular solution for \(\frac{dy}{dx} = 2x - 3y\) satisfying \(x=0\) and \(y=1\).

  1. The equation can be rewritten as

     \(\frac{dy}{dx} = 2x - 3y\)

     \(\implies \frac{dy}{dx} + 3y = 2x\)

     \(\implies \frac{dy}{dx} + P(x) \cdot y = Q(x)\)

     where \(P(x) = 3\) and \(Q(x) = 2x\)

  2. Then the integrating factor is \(I(x) = e^{\int P(x) dx} = e^{3x}\).

  3. Calculate \(\int Q(x) \cdot I(x) dx = \int 2x e^{3x} dx = \int 2x e^{3x} dx\).

     Clearly this integral should be calculated via integration by parts.

     Let \(u=2x\) and \(v' = e^{3x}\). Then \(u'=2\) and \(v = \frac{e^{3x}}{3}\).

     Then \(\int 2x \cdot e^{3x} dx = \int uv' dx = uv - \int u'vdx\)

     \(= 2x \cdot \frac{e^{3x}}{3} - \int 2 \cdot \frac{e^{3x}}{3} dx\)

     \(= \frac{2xe^{3x}}{3} - \frac{2}{3} \int e^{3x} dx\)

     \(= \frac{2xe^{3x}}{3} - \frac{2}{3} \cdot \frac{e^{3x}}{3} + C\)

     \(= \frac{2xe^{3x}}{3} - \frac{2e^{3x}}{9} + C\)

  4. The general solution is

     \(y = \frac{\int Q(x) \cdot I(x) dx}{I(x)}\)

     \(= \frac{\frac{2xe^{3x}}{3} - \frac{2e^{3x}}{9} + C}{e^{3x}}\)

     \(= \frac{2x}{3} - \frac{2}{9} + \frac{C}{e^{3x}}\)

  5. To find the particular solution for \(x=0, y=1\), we replace \(x\) and \(y\) with \(0\) and \(1\) respectively in the general solution, and get

     \(1 = \frac{2 \cdot 0}{3} - \frac{2}{9} + \frac{C}{e^{3 \cdot 0}}\)

     \(\implies 1 = - \frac{2}{9} + C\)

     \(\implies C = \frac{11}{9}\).

     Replace \(C\) with \(\frac{11}{9}\), and we get the particular solution for \(x=0, y=1\):

     \(y = \frac{2x}{3} - \frac{2}{9} + \frac{\frac{11}{9}}{e^{3x}} = \frac{2x}{3} - \frac{2}{9} + \frac{11}{9e^{3x}}\).

• Example 2
  

  Find the general solution for \(x^2 \frac{dy}{dx} + 3x = xy + x^4 e^{2x+1}\).

  1. The equation can be rewirtten as

     \(x^2 \frac{dy}{dx} + 3x = xy + x^4 e^{2x + 1}\)

     \(\implies x^2 \frac{dy}{dx} = -3x + xy + x^4 e^{2x + 1}\)

     \(\implies \frac{dy}{dx} = \frac{-3x + xy + x^4 e^{2x + 1}}{x^2}\)

     \(\implies \frac{dy}{dx} = \frac{-3}{x} + \frac{y}{x} + x^2 e^{2x + 1}\)

     \(\implies \frac{dy}{dx} - \frac{y}{x} = \frac{-3}{x} + x^2 e^{2x + 1}\)

     \(\implies \frac{dy}{dx} + P(x) \cdot y = Q(x)\)

     where \(P(x) = \frac{-1}{x}\) and \(Q(x) = \frac{-3}{x} + x^2 e^{2x + 1}\)

  2. The integrating factor is \(I(x) = e^{\int P(x) dx} = e^{\int \frac{-1}{x} dx} = e^{- \int \frac{1}{x} dx} = e^{-ln|x|}\).

     We have the formula \(e^{ln(f(x))} = f(x)\) because the natural log and the exponential operations are inverse to each other and thus can be offset. Can we simplify \(I(x) = e^{-ln|x|}\) using the formula? Yes, but we need first get rid of the negative sign before \(ln|x|\). By the property \(a lnb = ln(b^a)\), we have \(I(x) = e^{-ln|x|} = e^{ln(|x|^{-1})} = |x|^{-1} = \frac{1}{|x|}\).

     OK, it seems better, but there is an absolute value symbol in \(I(x)\), which may cause problems in Step 3 because we don't know how to integrate \(\int Q(x) I(x) dx\) if \(I(x)\) has an absolute value. For finding the solution to a linear first-order differential equation, we can directly remove the absolute value symbol in this kind of \(I(x)\). The reason is as follows. We know \(|x| = x\) if \(x>0\) and \(|x| = -x\) if \(x<0\), so \(I(x) = \frac{1}{x}\) if \(x>0\) and \(I(x) = \frac{1}{-x}\) if \(x<0\). If we discuss it case by case, then

     • For \(x>0\), you can find a general solution to the differential equation via Step 3 and Step 4 by using \(I(x) = \frac{1}{x}\);
     • For \(x<0\), you can find another general solution to the differential equation via Step 3 and Step 4 by using \(I(x) = \frac{1}{-x}\).

     You will see that these two solutions are exactly the same in form. That means, we can consider just the case \(x>0\) and let \(I(x) = \frac{1}{x}\).

     Hence, finally, we have removed the absolute value symbol and the integrating factor becomes \(I(x) = \frac{1}{x}\).

  3. Calculate \(\int Q(x) \cdot I(x) dx\).

     \(\int Q(x) \cdot I(x) dx = \int \left( \frac{-3}{x} + x^2 e^{2x + 1} \right) \cdot \frac{1}{x} dx\)

     \(= \int \frac{-3}{x^2} + x e^{2x + 1} dx\)

     \(= \int \frac{-3}{x^2} dx + \int x e^{2x + 1} dx\)

     \(= A + B\)

     where \(A = \int \frac{-3}{x^2} dx\) and \(B = \int x e^{2x + 1} dx\).

     Calculate \(A\):

     \(A = \int \frac{-3}{x^2} dx = (-3) \cdot \frac{x^{-2+1}}{-2+1} + C_1 = 3x^{-1} + C_1\)

     Calculate \(B\) by integration by parts:

     Let \(u = x\) and \(v' = e^{2x + 1}\). Then \(u' = 1\) and \(v = \frac{e^{2x + 1}}{2}\).

     \(B = \int x e^{2x + 1} dx\)

     \(= uv - \int u'v dx\)

     \(= x \cdot \frac{e^{2x + 1}}{2} - \int 1 \cdot \frac{e^{2x + 1}}{2} dx\)

     \(= \frac{xe^{2x + 1}}{2} - \frac{1}{2} \int e^{2x + 1} dx\)

     \(= \frac{xe^{2x + 1}}{2} - \frac{1}{2} \cdot \frac{e^{2x + 1}}{2} + C_2\)

     \(= \frac{xe^{2x + 1}}{2} - \frac{e^{2x + 1}}{4} + C_2\)

     Therefore, \(\int Q(x) \cdot I(x) dx = A + B = 3x^{-1} + \frac{xe^{2x + 1}}{2} - \frac{e^{2x + 1}}{4} + C\)

  4. The general solution is

     \(y = \frac{\int Q(x) \cdot I(x) dx}{I(x)}\)

     \(= \frac{3x^{-1} + \frac{xe^{2x + 1}}{2} - \frac{e^{2x + 1}}{4} + C}{\frac{1}{x}}\)

     \(= \left( 3x^{-1} + \frac{xe^{2x + 1}}{2} - \frac{e^{2x + 1}}{4} + C \right) \cdot x\)

     \(= 3 + \frac{x^2 e^{2x + 1}}{2} - \frac{xe^{2x + 1}}{4} + Cx\)

6.5.1 The equation

6.5.2 General Solution

• Formula
  

  The general solution to \(\frac{dy}{dx} = ry + D\) is

  \[ \boxed{ y = \frac{C e^{rx} - D}{r} } \]

• Derivation of The Formula
  

  The differential equation \(\frac{dy}{dx} = ry + D\)

  • is a separable differential equation because \(\frac{dy}{dx} = ry + D \implies \frac{dy}{dx} = \frac{1}{\frac{1}{ry + D}} = \frac{p(x)}{q(y)}\) where \(p(x) = 1\) and \(q(y) = \frac{1}{ry + D}\);
  • is also a linear first-order differential equation, because \(\frac{dy}{dx} = ry + D \implies \frac{dy}{dx} - ry = D \implies \frac{dy}{dx} + P(x)y = Q(x)\) where \(P(x) = -r\) and \(Q(x) = D\).

  Therefore, we can solve it by two different methods and of course they will give us the same answer.

  • Method 1 : Separable Differential Equation
    

    When we regard \(\frac{dy}{dx} = ry + D\) as a separable differential equation, the general solution is \(\int q(y) dy = \int p(x) dx\). From above discussion, here \(p(x) = 1\) and \(q(y) = \frac{1}{ry+D}\), so the general solution is \(\int \frac{1}{ry + D} dy = \int 1 dx\). Evaluate the two sides, and we get

    • \(LHS = \int \frac{1}{ry + D} dy\)

      Let \(u = ry+D\), so \(du = u'dy = rdy\). Then \(LHS = \int \frac{1}{ry+D} dy\) \(= \int \frac{1}{ry+D} \cdot \frac{r}{r} dy\) \(= \frac{1}{r} \int \frac{1}{ry+D} rdy\) \(= \frac{1}{r} \int \frac{1}{u} du\) \(= \frac{1}{r} \cdot ln|u| + C_1\) \(= \frac{1}{r} \cdot ln|ry+D| + C_1\) \(= \frac{1}{r} \cdot ln(ry+D) + C_1\) because \(ry+D>0\).

    • \(RHS = \int 1 dx = \int 1 dx = x + C_2\).

    Therefore, the general solution to \(\frac{dy}{dx} = ry + D\) is

    \(LHS = RHS\)

    \(\implies \frac{1}{r} \cdot ln(ry+D) + C_1 = x + C_2\)

    \(\implies \frac{1}{r} \cdot ln(ry+D) = x + C\)

    \(\implies ln(ry+D) = r(x + C)\)

    \(\implies ln(ry+D) = rx + C^*\) where \(C^* = rC\)

    \(\implies e^{ln(ry+D)} = e^{rx + C^*}\)

    \(\implies ry+D = e^{rx} \cdot e^{C^*}\)

    \(\implies ry+D = e^{rx} \cdot C^{**}\) where \(C^{**} = e^{C^*}\)

    \(\implies y= \frac{C^{**} e^{rx} - D}{r}\)

    or written as \[ y = \frac{C e^{rx} - D}{r} \]

  • Method 2 : Linear First-Order Differential Equation
    

    When we regard \(\frac{dy}{dx} = ry + D\) as a linear first-order differential equation, the general solution is \(y = \frac{\int Q(x) I(x) dx }{I(x)}\) where \(I(x)\) is the integrating factor.

    From above discussion, here \(P(x) = -r\) and \(Q(x) = D\), so the general solution is \(y = \frac{\int Q(x) I(x) dx }{I(x)}\).

    • \(I(x) = e^{\int P(x) dx} = e^{-rx}\)
    • \(\int Q(x)I(x) dx = \int D e^{-rx} dx = D \int e^{-rx} dx = D \cdot \frac{e^{-rx}}{-r} + C = \frac{-De^{-rx}}{r} + C\)

    • \(y = \frac{\int Q(x) I(x) dx }{I(x)}\) \(= \frac{\frac{-De^{-rx}}{r} + C}{e^{-rx}}\) \(= \frac{\frac{-De^{-rx}}{r}}{e^{-rx}} + \frac{C}{e^{-rx}}\) \(= \frac{-D}{r} + Ce^{rx}\) \(= \frac{-D}{r} + \frac{rCe^{rx}}{r}\) \(= \frac{-D}{r} + \frac{C^*e^{rx}}{r}\) \(= \frac{C^*e^{rx} - D}{r}\) where \(C^* = rC\)

      or written as \[ y = \frac{C e^{rx} - D}{r} \]

6.5.3 Particular Solution

6.5.4 Total Amount at a Specific Time

6.5.5 Example

7.1.1 Definition

• Example
  
  • \(3, 5, 7, 9, 11 \dots\) is a sequence.
  • Each number in \(3, 5, 7, 9, 11, \dots\) is called a term of the sequence.
  • We see
    • the 1st term is \(a_1 = 3 = 1 \cdot 2 + 1\)
    • the 2nd term is \(a_2 = 5 = 2 \cdot 2 + 1\)
    • the 3rd term is \(a_3 = 7 = 3 \cdot 2 + 1\)
    • the 4rd term is \(a_4 = 9 = 4 \cdot 2 + 1\)
    • the 5rd term is \(a_5 = 11 = 5 \cdot 2 + 1\)
    • \(\dots\)
    • the \(i\)th term is \(a_i = f(i) = i \cdot 2 + 1\)
    • so the general term of the sequence is \(a_i = f(i) = i \cdot 2 + 1\).

7.1.2 Sum of First \(n\) Terms of a Sequence

7.2.1 Definition

7.2.2 Convergence of Series

7.3.1 Introduction

7.3.2 Geometric Sequence

• Definition
  

  Suppose \(a_1, a_2, a_3, a_4, a_5, \dots\) is a sequence. If \(\frac{a_{i+1}}{a_i}\) equals the same constant \(r\) for all \(i = 1, 2, 3, \dots\), then the sequence is called a geometric sequence.

  • Example 1

    Suppose a sequence \(a_1 = 3, a_2 = 9, a_3 = 27, a_4 = 81, a_5 = 243, \dots\) is given. We see that

    • \(\frac{a_2}{a_1} = \frac{9}{3} = 3\)
    • \(\frac{a_3}{a_2} = \frac{27}{9} = 3\)
    • \(\frac{a_4}{a_3} = \frac{81}{27} = 3\)
    • \(\frac{a_5}{a_4} = \frac{243}{81} = 3\)
    • \(\dots\)

    That is, \(\frac{a_{i+1}}{a_i} = 3\) for all \(i = 1, 2, 3, \dots\), so this sequence is a geometric sequence.

  • Example 2

    Suppose a sequence \(a_1 = 3, a_2 = 9, a_3 = 21, \dots\) is given. We see that

    • \(\frac{a_2}{a_1} = \frac{9}{3} = 3\)
    • \(\frac{a_3}{a_2} = \frac{21}{9} = \frac{7}{3} \neq 3\)

    Since \(\frac{a_2}{a_1} \neq \frac{a_3}{a_2}\), this sequence is NOT a geometric sequence.

• Common Ratio
  

  For a geometric sequence \(a_1, a_2, a_3, a_4, a_5, \dots\), the constant \(r = \frac{a_{i+1}}{a_i}\) is called the common ratio.

  Remark:

  • \(r\) is called the common ratio because it is the same between any two consecutive terms.
  • For a geometric sequence, \(r = \frac{a_{i+1}}{a_i}\) is the same for all \(i = 1, 2, 3, \dots\), so the value of \(r\) can be cauculated simply by \(r = \frac{a_2}{a_1}\).
• General Term
  

  If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the general term of the sequence is \[ \boxed{a_i = a_1 r^{i-1}} \text{ for } i = 1,2,3,\dots .\]

  Remark: Clearly

  • \(a_2 = a_1 \cdot r = a_1 r^{2-1}\)
  • \(a_3 = a_2 \cdot r = (a_1 \cdot r) \cdot r = a_1 r^2 = a_1 r^{3-1}\)
  • \(a_4 = a_3 \cdot r = [a_2 \cdot r] \cdot r = [(a_1 \cdot r) \cdot r] \cdot r = a_1 r^3 = a_1 r^{4-1}\)
  • \(\dots\)
• Sum of First \(n\) Terms of a Geometric Sequence
  

  If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the sum of the first \(n\) terms, denoted by \(S_n\), is given by \[ \boxed{ S_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \dots + a_n = \frac{a_1(r^n - 1)}{r - 1} \text{ where } r\neq 1 } \]

7.3.3 Geometric Series

• Convergence of Geometric Series
  

  If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the corresponding geometric series \(\sum_{i=1}^{\infty} a_i\)

  • converges to \(\frac{a_1}{1-r}\) if \(r\in (-1, 1)\)
  • diverges if \(r\notin (-1,1)\)

7.3.4 Comprehensive Examples

• Example 1
  

  Suppose a geometric sequence \(2, -6, 18, -54, 162, \dots\) is given.

  • The first term is \(a_1 = 2\)
  • The common ratio is \(r = \frac{a_2}{a_1} = \frac{-6}{2} = -3\)
  • The general term is \(a_n = a_1 r^{n-1} = 2 \cdot (-3)^{n-1}\)
    • The term \(a_6 = a_1 r^{6-1} = 2 \cdot (-3)^5 = -486\)
    • The term \(a_7 = a_1 r^{7-1} = 2 \cdot (-3)^6 = 1458\)
  • The sum of first \(n\) terms is \(S_n = \frac{a_1(r^n - 1)}{r - 1}\) \(= \frac{2 \cdot ( (-3)^n - 1 )}{-3-1}\) \(= \frac{2 \cdot ( (-3)^n - 1 )}{-4}\) \(= \frac{(-3)^n - 1}{-2}\)
    • The sum of first \(2\) terms is \(S_2 = \frac{(-3)^2 - 1}{-2} = \frac{8}{-2} = -4\)
    • The sum of first \(5\) terms is \(S_5 = \frac{(-3)^5 - 1}{-2} = \frac{-244}{-2} = 122\)
  • Since the common ratio is \(-3 \notin (-1,1)\), the series \(\sum_{i=1}^{\infty} a_n\) diverges
• Example 2
  

  Suppose a geometric sequence \(2, \frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \frac{2}{81} \dots\) is given.

  • The first term is \(a_1 = 2\)
  • The common ratio is \(r\) \(= \frac{a_2}{a_1}\) \(= \frac{\frac{2}{3}}{2}\) \(= \frac{1}{3}\)
  • The general term is \(a_n = a_1 r^{n-1} = 2 \cdot \left( \frac{1}{3} \right)^{n-1}\)
    • The term \(a_6\) \(= a_1 r^{6-1}\) \(= 2 \cdot \left( \frac{1}{3} \right)^5\) \(= \frac{2}{243}\)
    • The term \(a_7\) \(= a_1 r^{7-1}\) \(= 2 \cdot \left( \frac{1}{3} \right)^6\) \(= \frac{2}{729}\)
  • The sum of first \(n\) terms is \(S_n\) \(= \frac{a_1(r^n - 1)}{r - 1}\) \(= \frac{2 \cdot \left[ \left( \frac{1}{3} \right)^n - 1 \right]}{\frac{1}{3} - 1}\) \(= \frac{2 \cdot \left[ \left( \frac{1}{3} \right)^n - 1 \right]}{\frac{-2}{3}}\) \(= -3 \left[ \left( \frac{1}{3} \right)^n - 1 \right]\) \(= -3 \cdot \left( \frac{1}{3} \right)^n + 3\) \(= - \left( \frac{1}{3} \right)^{n-1} + 3\)
    • The sum of first \(2\) terms is \(S_2 = - \left( \frac{1}{3} \right)^{2-1} + 3 = \frac{8}{3}\)
    • The sum of first \(5\) terms is \(S_5 = - \left( \frac{1}{3} \right)^{5-1} + 3 = \frac{242}{81}\)
  • Since the common ratio is \(\frac{1}{3} \in (-1,1)\), the series \(\sum_{i=1}^{\infty} a_n\) converges to \(\frac{a_1}{1-r} = \frac{2}{1-\frac{1}{3}} = 3\)
• Example 3
  

  Find \(\sum_{i=1}^{10} \frac{5^{3i}}{2^{i-1}}\).

  • Identify that this expression is the sum of first 10 terms of a sequence whose general term is \(a_i = \frac{5^{3i}}{2^{i-1}}\).

  • Show that the sequence is a geometric sequence. We have \(\frac{a_{i+1}}{a_i}\) \(= \frac{\frac{5^{3(i+1)}}{2^{(i+1)-1}}}{\frac{5^{3i}}{2^{i-1}}}\) \(= \frac{5^{3(i+1)}}{2^{(i+1)-1}} \cdot \frac{2^{i-1}}{5^{3i}}\) \(= \frac{2^{i-1}}{2^{(i+1)-1}} \cdot \frac{5^{3(i+1)}}{5^{3i}}\) \(= \frac{2^{i-1}}{2^i} \cdot \frac{5^{3i+3}}{5^{3i}}\) \(= \frac{1}{2} \cdot 5^3\) \(= \frac{125}{2}\).

    That is to say, \(\frac{a_{i+1}}{a_i} = \frac{125}{2}\) is a constant for all \(i=1,2,3,\dots\). By definition, the sequence is a geometric sequence.

  • The first term of the sequence is \(a_1 = \frac{5^{3\cdot 1}}{2^{1-1}} = 125\).
  • The common ratio of the sequence is \(r=\frac{125}{2}\).

  • Therefore, by formula, the sum of the first \(10\) terms of sequence \(a_i = \frac{5^{3i}}{2^{i-1}}\) is \(\sum_{i=1}^{10} \frac{5^{3i}}{2^{i-1}}\) \(= \frac{\left[\left(\frac{125}{2}\right)^{10} - 1\right]}{\frac{125}{2} - 1}\) \(= \frac{2\left[\left(\frac{125}{2}\right)^{10} - 1\right]}{123}\) \(= \frac{2\left(\frac{125}{2}\right)^{10} - 2}{123}\) \(= \frac{{125^{10}}{2^9} - 2}{123}\).

    You are not required to simplify it any more or write it as a decimal.

• Example 4
  

  Find the first \(6\) terms of the geometric sequence whose 2nd term is \(4\) and 5th term is \(32\).

  Basically, we should first find the general term of the sequence, and then calculate each one of the first \(6\) terms using the general term.

  By the general term of geometric sequences \(a_i = a_1 r^{i-1}\),

  • the 2nd term is \(a_2 = 4 = a_1 r^{2-1} = a_1 r \ \ (1)\)
  • the 5th term is \(a_5 = 32 = a_1 r^{5-1} = a_1 r^4 \ \ (2)\)

  Now we have to solve for \(a_1\) and \(r\) from the system consisting of equations \((1)\) and \((2)\). We use the substitution method.

  • From \((1)\), we get \(a_1 = \frac{4}{r} \ \ (3)\).
  • Put \((3)\) into \((2)\), and we get \(32 = \frac{4}{r} \cdot r^4 = 4r^3 \implies 8 = r^3 \implies r = 2 \ \ (4)\).
  • Put \((4)\) into \((3)\), and we get \(a_1 = \frac{4}{2} = 2\).

  Therefore, the general term of the sequence is \(a_i = a_1 r^{i-1} = 2 \cdot 2^{i-1} = 2^i\).

  In turn, we can calculate the first \(6\) terms

  • \(a_1 = 2^1 = 2\)
  • \(a_2 = 2^2 = 4\) or calculated by \(a_2 = a_1 r = 2 \cdot 2 = 4\)
  • \(a_3 = 2^3 = 8\) or calculated by \(a_3 = a_2 r = 4 \cdot 2 = 8\)
  • \(a_4 = 2^4 = 16\) or calculated by \(a_4 = a_3 r = 8 \cdot 2 = 16\)
  • \(a_5 = 2^5 = 32\) or calculated by \(a_5 = a_4 r = 16 \cdot 2 = 32\)
  • \(a_6 = 2^6 = 64\) or calculated by \(a_6 = a_5 r = 32 \cdot 2 = 64\)
• Example 5
  

  Check if the geometric series \(2 - \frac{4}{3} + \frac{8}{9} - \frac{16}{27} + \cdots\) converges or not. Find the sum if it converges.

  This is a geometric series corresponding to the geometric sequence \(2, \frac{-4}{3}, \frac{8}{9}, \frac{-16}{27}, \dots\). The 1st term of the sequence is \(a_1 = 2\) and the 2nd term of the sequence is \(a_2 = \frac{-4}{3}\), so the common ratio of the sequence is \(r = \frac{a_2}{a_1} = \frac{\frac{-4}{3}}{2} = \frac{-2}{3} \in (-1,1)\).

  Hence, by the rule, the series converges to \(\frac{a_1}{1-r} = \frac{2}{1-\frac{-2}{3}} = \frac{6}{5}\).

• Example 6
  

  Check if the geometric series \(\sum_{i=1}^{\infty} \frac{5^{3i}}{2^{i-1}}\) converges or not. Find the sum if it converges.

  This is a geometric series corresponding to the geometirc sequence whose general term is \(a_i = \frac{5^{3i}}{2^{i-1}}\). The 1st term of the sequence is \(a_1 = \frac{5^{3\cdot 1}}{2^{1-1}} = 125\) and the 2nd term of sequence is \(\frac{5^{3\cdot 2}}{2^{2-1}} = \frac{15625}{2}\), so the common ratio of the sequence is \(r = \frac{a_2}{a_1} = \frac{\frac{15625}{2}}{125} = \frac{125}{2} \notin (-1,1)\).

  Hence, by the rule, the series diverges.

• Example 7
  

  Express the repeated decimal \(0.535353\dots\) as a rational number.

  A number is rational if it can be written as a fraction between two integers, so the nature of the problem is to convert \(0.535353\dots\) into a fraction of two integers.

  We note that \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) is a series corresponding to the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\).

  Now we verify that the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\) is a geometric sequence : we see that \(\frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \cdots = \frac{a_{i+1}}{a_i} = 0.01\) is a constant for any \(i = 1, 2, 3, \dots\), so the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\) is a geometric sequence with common ratio \(r = 0.01 \in (-1, 1)\).

  Therefore, \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) is the geometric series corresponding to the geometric sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\).

  By the rule, sine \(r \in (-1,1)\), the geometric series \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) converges to \(\frac{a_1}{1-r} = \frac{0.53}{1-0.01} = \frac{0.53}{0.99} = \frac{53}{99}\).

  Finally, we have \(0.535353\dots = \frac{53}{99}\), which is a fraction between two integers, so it is a rational number.

7.4.1 Indeterminate Form

7.4.2 L'Hospital's Rule

7.4.3 Remark

• Check the condition
  

  Before applying L'Hospital's Rule, we should ALWAYS check whether the condition "\(\lim\limits_{x\to a} f(x) = 0\) and \(\lim\limits_{x\to a} g(x) = 0\) hold at the same time" or "\(\lim\limits_{x\to a} f(x) = \pm \infty\) and \(\lim\limits_{x\to a} g(x) = \pm\infty\) hold at the same time" holds. Otherwise, the rule does not apply.

• Repeated Application
  

  Sometimes, after the first application of L'Hospital's Rule, the limit is still an indeterminate form. In this case, we may apply the L'Hospital's Rule for another time to see whether the problem gets simpler.

7.4.4 Examples

• Example 1 : L'Hospital's Rule Does NOT Apply
  

  Find \(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 - x - 6}\).

  Here we identify

  • \(f(x) = x^2 - 4\)
  • \(g(x) = x^2 - x -6\)

  We know

  • \(\lim\limits_{x\to 2} f(x) = \lim\limits_{x\to 2} x^2 - 4 = 2^2 - 4 = 0\)
  • \(\lim\limits_{x\to 2} g(x) = \lim\limits_{x\to 2} x^2 - x - 6 = 2^2 - 2 - 6 = -4 \neq 0\)

  so the L'Hospital's Rule doesn't work for this problem.

  As we learned in the Limit chapter, this limit can be found just by substitution:

  \(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 - x - 6} = \frac{0}{-4} = 0\).

• Example 2 : \(0/0\) indeterminate form
  

  Find \(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 + x - 6}\).

  Here we identify

  • \(f(x) = x^2 - 4\)
  • \(g(x) = x^2 + x -6\)

  We know

  • \(\lim\limits_{x\to 2} f(x) = \lim\limits_{x\to 2} x^2 - 4 = 2^2 - 4 = 0\)
  • \(\lim\limits_{x\to 2} g(x) = \lim\limits_{x\to 2} x^2 + x - 6 = 2^2 + 2 - 6 = 0\)

  so the limit is an indeterminate form.

  By L'Hospital's Rule,

  \(\lim\limits_{x\to 2} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to 2} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to 2} \frac{2x}{2x+1}\) \(= \frac{2 \cdot 2}{2\cdot 2 + 1}\) \(= \frac{4}{5}\)

• Example 3 : \(\infty/\infty\) indeterminate form, and repeated application of L'Hospital's Rule
  

  Find \(\lim\limits_{x\to \infty} \frac{x^2 - 4}{x^2 - x - 6}\).

  Here we identify

  • \(f(x) = x^2 - 4\)
  • \(g(x) = x^2 - x -6\)

  We know

  • \(\lim\limits_{x\to \infty} f(x)\) \(= \lim\limits_{x\to \infty} x^2 - 4\) \(= \infty^2 - 4 = \infty\)

  • \(\lim\limits_{x\to \infty} g(x)\) \(= \lim\limits_{x\to \infty} x^2 - x - 6\) \(= \lim\limits_{x\to \infty} x (x - 1 - \frac{6}{x})\) \(= \lim\limits_{x\to \infty} x \cdot \lim\limits_{x\to \infty}x - 1 - \frac{6}{x}\) \(= \infty \cdot (\infty - 1 - 0)\) \(= \infty \cdot \infty = \infty\).

    (Note : for the denominator, if we directly substitute \(\infty\) for \(x\) in \(x^2 - x - 6\), then we get \(\infty^2 - \infty - 6 = \infty - \infty - 6\) and we don't know how to evaluate the indeterminate difference \(\infty - \infty\). We don't know whether it exists or not, and we don't know whether it is \(0\), \(\infty\) or any other constant. This is the reason why we do the factorization \(x^2 - x - 6 = x (x - 1 - \frac{6}{x})\) and then find \(\lim\limits_{x\to \infty} x^2 - x - 6 = \infty\).)

  The limit is an indeterminate form \(\infty/\infty\).

  By L'Hospital's Rule,

  \(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{2x}{2x+1}\) \(= \frac{2 \cdot \infty}{2\cdot \infty + 1}\) \(= \frac{\infty}{\infty}\)

  which is again an indeterminate form \(\infty/\infty\).

  Therefore, let's apply the L'Hospital's Rule for another time and get

  \(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{2x}{2x+1}\) \(= \lim\limits_{x\to \infty} \frac{(2x)'}{(2x+1)'}\) \(= \lim\limits_{x\to \infty} \frac{2}{2}\) \(= \lim\limits_{x\to \infty} 1\) \(= 1\).

  Recall that we actually could find this limit without using L'Hospital's Rule. We can just divide both numerator and denominator by \(x^2\) and get:

  \(\lim\limits_{x\to \infty} \frac{x^2 - 4}{x^2 - x - 6}\) \(= \lim\limits_{x\to \infty} \frac{\frac{x^2 - 4}{x^2}}{\frac{x^2 - x - 6}{x^2}}\) \(= \lim\limits_{x\to \infty} \frac{1 - \frac{4}{x^2}}{1 - \frac{1}{x} - \frac{6}{x^2}}\) \(= \frac{1 - \frac{4}{\infty^2}}{1 - \frac{1}{\infty} - \frac{6}{\infty^2}}\) \(= \frac{1 - 0}{1 - 0 - 0}\) \(= 1\).

• Example 4 : L'Hospital's Rule applicable but not helpful
  

  Find \(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\).

  Here we identify

  • \(f(x) = \sqrt{x+1}\)
  • \(g(x) = \sqrt{x-1}\)

  We know

  • \(\lim\limits_{x\to \infty} f(x) = \lim\limits_{x\to \infty} \sqrt{x+1} = \sqrt{\infty+1} = \infty\)
  • \(\lim\limits_{x\to \infty} g(x) = \lim\limits_{x\to \infty} \sqrt{x-1} = \sqrt{\infty-1} = \infty\)

  The limit is an indeterminate form \(\infty/\infty\).

  By L'Hospital's Rule,

  \(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{\frac{1}{2\sqrt{x+1}}}{\frac{1}{2\sqrt{x-1}}}\) \(= \lim\limits_{x\to \infty} \frac{1}{2\sqrt{x+1}} \cdot 2\sqrt{x-1} = \lim\limits_{x\to \infty} \frac{\sqrt{x-1}}{\sqrt{x+1}}\)

  which is again an indeterminate form \(\infty/\infty\).

  Should we apply the L'Hospital's Rule for one more time? Let's try it:

  \(\lim\limits_{x\to \infty} \frac{\sqrt{x-1}}{\sqrt{x+1}}\) \(= \lim\limits_{x\to \infty} \frac{\frac{1}{2\sqrt{x-1}}}{\frac{1}{2\sqrt{x+1}}}\) \(= \lim\limits_{x\to \infty} \frac{1}{2\sqrt{x-1}} \cdot 2\sqrt{x+1} = \lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\)

  which is exactly the original problem!

  As you can imagine, even if we apply the L'Hospital's Rule for even more times, the problem still doesn't get simpler.

  Therefore, although L'Hospital's Rule is applicable here, it doesn't directly help us solve the problem. In this case, we need first make some change to the original problem.

  We put the quantities under square root symbols together by

  \(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\) \(=\lim\limits_{x\to \infty} \sqrt{\frac{x+1}{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} \frac{x+1}{x-1}}\)

  Now we can find \(\lim\limits_{x\to \infty} \frac{x+1}{x-1}\) by

  • L'Hospital's Rule, ie, \(\lim\limits_{x\to \infty} \frac{x+1}{x-1} = \lim\limits_{x\to \infty} \frac{1}{1} = 1\), or
  • dividing both numerator and demoninator by \(x\), ie, \(\lim\limits_{x\to \infty} \frac{x+1}{x-1}\) \(= \lim\limits_{x\to \infty} \frac{\frac{x+1}{x}} {\frac{x-1}{x}}\) \(= \lim\limits_{x\to \infty} \frac{1 + \frac{1}{x}} {1 - \frac{1}{x}}\) \(= \frac{1 + 0} {1 - 0}\) \(= 1\).

  Finally, the original limit

  \(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} \frac{x+1}{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} 1}\) \(=1\).