Integration using Substitution: Learn It 2

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Integration using Substitution: Learn It 2

Substitution for 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{array}{cc} \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{array}

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

and we have the desired result.

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

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 $\frac{1}{6}.$

Then,

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

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

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 this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.5 Substitution” here (opens in new window).

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

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} \int_{0}^{1} x e^{4 x^{2} + 3} d x & = \frac{1}{8} \int_{3}^{7} e^{u} d u \\ = \frac{1}{8} e^{u} |_{3}^{7} \\ = \frac{e^{7} - e^{3}}{8} \\ \approx 134.568. \end{array}

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

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.

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

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

\int_{0}^{π / 2} \cos^{2} θ d θ = \int_{0}^{π / 2} \frac{1 + \cos 2 θ}{2} d θ .

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

Then,

\begin{array}{cc} \int_{0}^{π / 2} (\frac{1 + \cos 2 θ}{2}) d θ & = \int_{0}^{π / 2} (\frac{1}{2} + \frac{1}{2} \cos 2 θ) d θ \\ = \frac{1}{2} \int_{0}^{π / 2} d θ + \frac{1}{2} \int_{0}^{π / 2} \cos 2 θ d θ . \end{array}

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

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{matrix} \frac{1}{2} \int_{0}^{π / 2} d θ + \frac{1}{2} \int_{0}^{π / 2} \cos 2 θ d θ & = \frac{1}{2} \int_{0}^{π / 2} d θ + \frac{1}{2} (\frac{1}{2}) \int_{0}^{π} \cos u d u \\ = \frac{θ}{2} |_{θ = 0}^{θ = π / 2} + \frac{1}{4} \sin u |_{u = 0}^{u = θ} \\ = (\frac{π}{4} - 0) + (0 - 0) = \frac{π}{4} . \end{matrix}

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

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.5 Substitution” here (opens in new window).