Substitution for Definite Integrals

Learning Outcomes

Use substitution to evaluate definite integrals.

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

Substitution with Definite Integrals

Let $u=g(x)$ and let ${g}^{\text{′}}$ be continuous over an interval $\left[a,b\right],$ and let $f$ be continuous over the range of $u=g(x).$ Then,

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u

${\displaystyle\int }_{a}^{b}f(g(x)){g}^{\prime }(x)dx={\displaystyle\int }_{g(a)}^{g(b)}f(u)du$

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $F(x)$ is an antiderivative of $f(x),$ we have

\int f (g (x)) g^{'} (x) d x = F (g (x)) + C

$\displaystyle\int f(g(x)){g}^{\prime }(x)dx=F(g(x))+C$

Then

\begin{matrix} \int_{a}^{b} f [g (x)] g^{'} (x) d x & = {F (g (x)) |}_{x = a}^{x = b} = F (g (b)) - F (g (a)) = {F (u) |}_{u = g (a)}^{u = g (b)} = \int_{g (a)}^{g (b)} f (u) d u, \end{matrix}

$\begin{array}{cc}{\displaystyle\int }_{a}^{b}f\left[g(x)\right]{g}^{\prime }(x)dx\hfill & ={F(g(x))|}_{x=a}^{x=b}\hfill \\ & =F(g(b))-F(g(a))\hfill \\ & ={F(u)|}_{u=g(a)}^{u=g(b)}\hfill \\ \\ \\ & ={\displaystyle\int }_{g(a)}^{g(b)}f(u)du,\hfill \end{array}$

and we have the desired result.

example: Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate ${\displaystyle\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.$

Show Solution

Let $u=1+2{x}^{3},$ so $du=6{x}^{2}dx.$ Since the original function includes one factor of $x$ ² and $du=6{x}^{2}dx,$ multiply both sides of the du equation by $1\text{/}6.$ Then,

\begin{matrix} d u & = & 6 x^{2} d x \frac{1}{6} d u & = & x^{2} d x . \end{matrix}

$\begin{array}{ccc}du\hfill & =\hfill & 6{x}^{2}dx\hfill \\ \frac{1}{6}du\hfill & =\hfill & {x}^{2}dx.\hfill \end{array}$

To adjust the limits of integration, note that when $x=0,u=1+2(0)=1,$ and when $x=1,u=1+2(1)=3.$ Then

\int_{0}^{1} x^{2} {(1 + 2 x^{3})}^{5} d x = \frac{1}{6} \int_{1}^{3} u^{5} d u .

${\displaystyle\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx=\frac{1}{6}{\displaystyle\int }_{1}^{3}{u}^{5}du.$

Evaluating this expression, we get

\begin{matrix} \frac{1}{6} \int_{1}^{3} u^{5} d u & = (\frac{1}{6}) (\frac{u^{6}}{6}) |_{1}^{3} = \frac{1}{36} [{(3)}^{6} - {(1)}^{6}] = \frac{182}{9} . \end{matrix}

$\begin{array}{}\\ \\ \frac{1}{6}{\displaystyle\int }_{1}^{3}{u}^{5}du\hfill & =(\frac{1}{6})(\frac{{u}^{6}}{6}){|}_{1}^{3}\hfill \\ & =\frac{1}{36}\left[{(3)}^{6}-{(1)}^{6}\right]\hfill \\ & =\frac{182}{9}.\hfill \end{array}$

Watch the following video to see the worked solution to Example: Using Substitution to Evaluate a Definite Integral.

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Try It

Use substitution to evaluate the definite integral ${\displaystyle\int }_{-1}^{0}y{(2{y}^{2}-3)}^{5}dy.$

Show Solution

$\frac{91}{3}$

example: Using Substitution with an Exponential Function

Use substitution to evaluate ${\displaystyle\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.$

Show Solution

Let $u=4{x}^{3}+3.$ Then, $du=8xdx.$ To adjust the limits of integration, we note that when $x=0,u=3,$ and when $x=1,u=7.$ So our substitution gives

$\begin{array}{cc}{\displaystyle\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\hfill & =\frac{1}{8}{\displaystyle\int }_{3}^{7}{e}^{u}du\hfill \\ \\ & =\frac{1}{8}{e}^{u}{|}_{3}^{7}\hfill \\ & =\frac{{e}^{7}-{e}^{3}}{8}\hfill \\ & \approx 134.568.\hfill \end{array}$

Try It

Use substitution to evaluate ${\displaystyle\int }_{0}^{1}{x}^{2} \cos \left(\frac{\pi }{2}{x}^{3}\right)dx.$

Show Solution

$\frac{2}{3\pi }\approx 0.2122$

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for $u$ after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in the following examples.

example: Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate ${\displaystyle\int }_{0}^{\pi \text{/}2}{ \cos }^{2}\theta d\theta .$

Show Solution

Let us first use a trigonometric identity to rewrite the integral. The trig identity ${ \cos }^{2}\theta =\frac{1+ \cos 2\theta }{2}$ allows us to rewrite the integral as

${\displaystyle\int }_{0}^{\pi \text{/}2}{ \cos }^{2}\theta d\theta ={\displaystyle\int }_{0}^{\pi \text{/}2}\frac{1+ \cos 2\theta }{2}d\theta .$

Then,

$\begin{array}{cc}{\displaystyle\int }_{0}^{\pi \text{/}2}\left(\frac{1+ \cos 2\theta }{2}\right)d\theta \hfill & ={\displaystyle\int }_{0}^{\pi \text{/}2}\left(\frac{1}{2}+\frac{1}{2} \cos 2\theta \right)d\theta \hfill \\ \\ \\ & =\frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}2}d\theta + \frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}2} \cos 2\theta d\theta .\hfill \end{array}$

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $u=2\theta .$ Then, $du=2d\theta ,$ or $\frac{1}{2}du=d\theta .$ Also, when $\theta =0,u=0,$ and when $\theta =\pi \text{/}2,u=\pi .$ Expressing the second integral in terms of $u$ , we have

$\begin{array}{}\\ \\ \frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}2} \cos 2\theta d\theta \hfill & =\frac{1}{2}{\displaystyle\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}(\frac{1}{2}){\displaystyle\int }_{0}^{\pi } \cos udu\hfill \\ & =\frac{\theta }{2}{|}_{\theta =0}^{\theta =\pi \text{/}2}+\frac{1}{4} \sin u{|}_{u=0}^{u=\theta }\hfill \\ & =(\frac{\pi }{4}-0)+(0-0)=\frac{\pi }{4}.\hfill \end{array}$

Watch the following video to see the worked solution to Example: Using Substitution to Evaluate a Trigonometric Integral.

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Module 5: Integration