5.6: Substitution with Definite Integrals

Learning Objectives

• Use substitution to evaluate definite integrals.

Substitution with Definite Integrals

Let $u=g(x)$ and let $g'$ be continuous over an interval $[a,b]$, and let $f$ be continuous over the range of $u=g(x).$ Then,

$$∫^b_af(g(x))g′(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du. \nonumber$$

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

$$∫f(g(x))g′(x)\,dx=F(g(x))+C. \nonumber$$

Then

$$\begin{align*} ∫^b_af[g(x)]g′(x)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \\[4pt] &=F(g(b))−F(g(a)) \\[4pt] &= F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[4pt] &=∫^{g(b)}_{g(a)}f(u)\,du \end{align*}$$

and we have the desired result.

Example $\PageIndex{1}$: Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate $$∫^1_0x^2(1+2x^3)^5\,dx. \nonumber$$

Solution

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

$$\begin{align*} du &=6x^2\,dx \\[4pt] \text{becomes}\quad \dfrac{1}{6}du &=x^2\,dx. \end{align*}$$

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

$$∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du. \nonumber$$

Evaluating this expression, we get

$$\begin{align*} \dfrac{1}{6}∫^3_1u^5\,du &=\left(\dfrac{1}{6}\right)\left(\dfrac{u^6}{6}\right)\Big|^3_1 \\[4pt] &=\dfrac{1}{36}\big[(3)^6−(1)^6\big] \\[4pt] &=\dfrac{182}{9}. \end{align*}$$

Exercise $\PageIndex{1}$

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

Hint

Use the steps from Example $\PageIndex{5}$ to solve the problem.

Answer

$\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy = \dfrac{91}{3}$

Exercise $\PageIndex{2}$

Use substitution to evaluate $\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx.$

Hint

Use the process from Example $\PageIndex{5}$ to solve the problem.

Answer

$\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx = \dfrac{2}{3π}≈0.2122$

Example $\PageIndex{2}$: Using Substitution with an Exponential Function

Use substitution to evaluate $$∫^1_0xe^{4x^2+3}\,dx. \nonumber$$

Solution

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

$$\begin{align*} ∫^1_0xe^{4x^2+3}\,dx &= \dfrac{1}{8}∫^7_3e^u\,du \\[4pt] &=\dfrac{1}{8}e^u\Big|^7_3 \\[4pt] &=\dfrac{e^7−e^3}{8} \\[4pt] &≈134.568 \end{align*}$$

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 Example $\PageIndex{7}$.

Example $\PageIndex{3}$: Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate $$∫^{π/2}_0\cos^2θ\,dθ. \nonumber$$

Solution

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

$$∫^{π/2}_0\cos^2θ\,dθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ. \nonumber$$

Then,

$$\begin{align*} ∫^{π/2}_0\left(\dfrac{1+\cos2θ}{2}\right)\,dθ &=∫^{π/2}_0\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2θ\right)\,dθ \\[4pt] &=\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ. \end{align*}$$

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

$$\begin{align*}\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0 \cos 2θ\,dθ &=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}\left(\dfrac{1}{2}\right)∫^π_0 \cos u \,du \\[4pt] &=\dfrac{θ}{2}\,\bigg|^{θ=π/2}_{θ=0}+\dfrac{1}{4}\sin u\,\bigg|^{u=θ}_{u=0} \\[4pt] &=\left(\dfrac{π}{4}−0\right)+(0−0)=\dfrac{π}{4} \end{align*}$$