Substitution for Definite Integrals

Changing the Limits of Integration via Substitution

When we evaluate the definite integral \(\int_a^b f(x) ~dx\), 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 apply the substitution \(u = g(x)\), then we have to rewrite the limits of integration in terms of the new variable \(u\). The substitution rule for definite integrals tells us exactly how to do that.

Substitution Rule for Definite Integrals

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

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

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 using the substitution \(u = 2x^3+1\).

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

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.

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

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 using the substitution \(u = -3x/2\).

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

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 property of definite integrals listed on page 3. 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 using the substitution \(u = 3x^2-2\).

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

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.

\[\begin{align*} \frac{1}{6}\int_{u=1}^{u=25} u^{1/2} ~du &= \frac{1}{6}\cdot\frac{2}{3}u^{3/2}\Biggr|_1^{25} && \hbox{Power rule}\\ &= \frac{1}{9}u^{3/2}\Biggr|_1^{25} && \hbox{Simplify}\\ &= \frac{1}{9}(25^{3/2} - 1^{3/2}) && \hbox{Plug in limits of integration}\\ &= \frac{1}{9}(5^{3} - 1) && \hbox{Since $25^{3/2} = (25^{1/2})^3 = 5^3$}\\ &= 124/9 \end{align*}\]

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 using the substitution \(u = 1+e^x\).

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

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.

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

Therefore,

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