Substitution for Definite Integrals

Changing the Limits of Integration via Substitution

Substitution Rule for Definite Integrals

When evaluating the definite integral \(\displaystyle \int_a^b f(x) ~dx\), the limits of integration are assumed to be in terms of the variable \(x\) (i.e., we are integrating from \(x = a\) to \(x = b\)). If we change the variable by applying the substitution \(u = g(x)\), then we also need to change the limits of integration so that they are in terms of the new variable \(u\). The substitution rule for definite integrals, as stated below, tells us exactly how to do that.

\[\int_{x=a}^{x=b} f(g(x))g'(x)~dx = \int_{u=g(a)}^{u=g(b)} f(u) ~du\]

where \(u=g(x)\) and \(du = g'(x) ~dx\).

Method of Integration by Substitution for Definite Integrals

1. Write the given integral in one of the following two forms:

   \[\int_{x=a}^{x=b} [g(x)]^n g'(x) ~dx \qquad \text{or} \qquad \int_{x=a}^{x=b} e^{g(x)} g'(x) ~dx\]

2. Let \(u=g(x)\) and compute the corresponding differential \(du = g'(x) dx\).

3. Rewrite the integral in terms of \(u\) and \(du\). This includes changing the limits of integration so that the new lower limit of integration is \(g(a)\) and the new upper limit of integration is \(g(b)\).

4. Evaluate the integral in terms of the variable \(u\).

5. Plug in the new limits of integration, \(g(b)\) and \(g(a)\).

Example 1

Using substitution to integrate rational functions

Evaluate \(\displaystyle \int_{1}^{2} \frac{6x^2}{2x^3 + 1}~dx\).

Step 1:   Identify a suitable substitution.

Based on rewriting the integral in the following form

\[\int_{x=1}^{x=2} (2x^3+1)^{-1} 6x^2 ~dx\]

let \(u = 2x^3+1\) and \(du = 6x^2 ~du\).

Step 2:   Determine the new limits of integration.

If \(u = 2x^3+1\), then

\[\begin{align*} \text{when } x=2 ~~&\Longrightarrow~~ u = 2(2)^3 + 1 = 17 \\ \text{when } x=1 ~~&\Longrightarrow~~ u = 2(1)^3 + 1 = 3 \end{align*}\]

Step 3:   Rewrite the integral in terms of \(u\) and \(du\).

\[\int_{x=1}^{x=2} (2x^3+1)^{-1} 6x^2 ~dx = \int_{u=3}^{u=17} u^{-1} ~du\]

Step 4:   Evaluate the integral.

\[\begin{align*} \int_{u=3}^{u=17} \frac{1}{u}~du &= \ln|u| \Big|_3^{17} && \text{since $\ln|u|$ is an antiderivative of $\dfrac{1}{u}$}\\ &= \ln(17) - \ln(3) && \text{plug in limits of integration}\\ &= \ln(17/3) && \text{since $\ln(m/n) = \ln(m) - \ln(n)$} \end{align*}\]

Therefore,

\[\int_{1}^{2} \frac{6x^2}{2x^3 + 1}~dx = \ln(17/3).\]

Example 2

Finding the area under a graph

Find the area of the region under the graph of \(f(x) = e^{-3x/2}\) between \(x = -2\) and \(x = 6\).

Step 1:   Find the definite integral that corresponds to the area of the given region.

Since \(e^{-3x/2} \geq 0\) for all \(x\), the area under the graph of \(f\) on \([-2,6]\) is given by

\[ \int_{x=-2}^{x=6} e^{-3x/2} ~dx.\]

Step 2:   Identify a suitable substitution.

Let \(u = -3x/2\) and \(du = -\dfrac{3}{2}~dx\), or equivalently \(-\dfrac{2}{3} ~du = dx\).

Step 3:   Determine the new limits of integration.

If \(u = -3x/2\), then

\[\begin{align*} \text{when } x=6 ~~&\Longrightarrow~~ u = -3(6)/2 = -9 \\ \text{when } x=-2 ~~&\Longrightarrow~~ u = -3(-2)/2 = 3 \end{align*}\]

Step 4:   Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} \int_{x=-2}^{x=6} e^{-3x/2} ~dx &= \int_{u=3}^{u=-9} e^{u} \left(-\frac{2}{3}\right)~du \\ &= -\frac{2}{3}\int_{u=3}^{u=-9} e^{u} ~du \end{align*}\]

Notice that the new lower limit of integration (\(u=3\)) is larger than the new upper limit (\(u=-9\)). There is nothing wrong with this and we will continue the calculation in the next step as is. However, it is possible to switch limits of integration using the second Properties of the Definite Integral. Specifically,

\[ -\frac{2}{3}\int_{u=3}^{u=-9} e^{u} ~du = \frac{2}{3}\int_{u=-9}^{u=3} e^{u} ~du.\]

Step 5:   Evaluate the integral.

\[\begin{align*} -\frac{2}{3}\int_{u=3}^{u=-9} e^{u} ~du &= -\frac{2}{3} e^{u} \Biggr|_{u=3}^{u=-9} \\ &= -\frac{2}{3}(e^{-9} - e^3) \\ &= \frac{2}{3}(e^{3} - e^{-9}) \end{align*}\]

Therefore, the area under the graph of \(f\) on \([-2,6]\) is given by \(\dfrac{2}{3}\left(e^3 - e^{-9}\right)\).

Observation

The integral in Step 1 can be evaluated without using substitution. Try using the following formula instead.

\[ \int e^{ax} ~dx = \frac{1}{a}e^{ax} + C ~~~~~ \hbox{for } a\neq 0.\]

Example 3

Using substitution

Evaluate \(\displaystyle \int_{1}^{3} x\sqrt{3x^2-2}~dx\).

Step 1:   Identify a suitable substitution.

Based on rewriting the integral in the following form

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

let \(u = 3x^2-2\) and \(du = 6x ~dx\), or equivalently \(\dfrac{1}{6} du = x ~dx\).

Step 2:   Determine the new limits of integration.

If \(u = 3x^2-2\), then

\[\begin{align*} \text{when } x=3 ~~&\Longrightarrow~~ u = 3(3)^2 - 2 = 25 \\ \text{when } x=1 ~~&\Longrightarrow~~ u = 3(1)^2 - 2 = 1 \end{align*}\]

Step 3:   Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} \int_{x=1}^{x=3} (3x^2-2)^{1/2} ~x ~dx &= \int_{u=1}^{u=25} u^{1/2} \frac{1}{6}~du\\ &= \frac{1}{6}\int_{u=1}^{u=25} u^{1/2} ~du \end{align*}\]

Step 4:   Evaluate the integral.

\[\begin{align*} \frac{1}{6}\int_{u=1}^{u=25} u^{1/2} ~du &= \frac{1}{6}\cdot\frac{2}{3}u^{3/2}\Biggr|_1^{25} && \text{power rule}\\ &= \frac{1}{9}u^{3/2}\Biggr|_1^{25} && \text{simplify}\\ &= \frac{1}{9}(25^{3/2} - 1^{3/2}) && \text{plug in limits of integration}\\ &= \frac{1}{9}(5^{3} - 1) && \text{since $25^{3/2} = (25^{1/2})^3 = 5^3$}\\ &= 124/9 \end{align*}\]

Therefore,

\[\int_{1}^{3} x\sqrt{3x^2-2}~dx = 124/9.\]

Example 4

Using substitution

Evaluate \(\displaystyle \int_{0}^{2} \frac{e^x}{1+e^x} ~dx\).

Step 1:   Identify a suitable substitution.

Based on rewriting the integral in the following form

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

let \(u = 1+e^x\) and \(du = e^x ~dx\).

Step 2:   Determine the new limits of integration.

If \(u = 1+e^x\), then

\[\begin{align*} \text{when } x=2 ~~&\Longrightarrow~~ u = 1+e^2 \\ \text{when } x=0 ~~&\Longrightarrow~~ u = 1+e^0 = 2 \end{align*}\]

Step 3:   Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} \int_{x=0}^{x=2} (1+e^x)^{-1} ~e^x ~dx &= \int_{u=2}^{u=1+e^2} u^{-1} ~du\\ &= \int_{u=2}^{u=1+e^2} \frac{1}{u} ~du \end{align*}\]

Step 4:   Evaluate the integral.

\[\begin{align*} \int_{u=2}^{u=1+e^2} \frac{1}{u} ~du &= \ln|u| \Big|_2^{1+e^2} \\ &= \ln(1+e^2)-\ln(2) && \text{plug in limits of integration}\\ &= \ln\left(\frac{1+e^2}{2}\right)&& \text{since $\ln(m/n) = \ln(m) - \ln(n)$} \end{align*}\]

Therefore,

\[\int_{0}^{2} \frac{e^x}{1+e^x} ~dx = \ln\left(\frac{1+e^2}{2}\right).\]