Integral

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

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 \(f(x) = 2x\).

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 \(G(x)\).

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 \(\int f(x) dx\).

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 \(\int f(x) dx\).

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 \(\int f(x) dx\).

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 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 \(u\).

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.

• 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

• 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,

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

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

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

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

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

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

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

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

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

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

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 \[ \int (u \cdot v') dx = u \cdot v - \int (u' \cdot v) dx \]

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.

• 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

  \(\int ln(x) dx\)

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

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

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

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

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 \(\int xe^{x^2} dx\) \(= \frac{1}{2} \int e^{x^2} 2x dx\) \(= \frac{1}{2} \int e^u du\) \(= \frac{1}{2} \cdot e^u + C\) \(= \frac{1}{2} \cdot e^{x^2} + C\).

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.

• 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.

• "\(dx\)" 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.

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,

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

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

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

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

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

\(= 21\)

• 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.

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.

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.

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.

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.

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.

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

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

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

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

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

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

\(= 100\)

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 :

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

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

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

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

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

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

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

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

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

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