U-Substitution of Definite Integrals

So we have looked at a method for evaluating integrals using the U-substitution technique, however, all of the examples thus far have been indefinite integrals. The technique is similar for definite integrals, however, there is an extra step that we must always following regarding the lower and upper bounds of the definite integral.

For example, consider the following definite integral $\int_{1}^{4} \frac{2x}{x^2 + 1} \: dx$.

First, let $u = x^2 + 1$ so that $du = 2x \: dx$. Making this substitution, we get:

(1)

\begin{align} \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \int_{\alpha}^{\beta} \frac{1}{u} \: du \end{align}

Note that the upper and lower bounds of integration are not necessarily the same (hence we used $\alpha$ and $\beta$ as arbitrary placeholders). We can calculate them though using our substitution equation. Recall that $u = x^2 + 1$. Let's evaluate $u$ at $x = 1$, and at $x = 4$ as follows:

(2)

\begin{align} \alpha = u \: \rvert_{x=1} = 1^2 + 1 = 2 \\ \beta = u \: \rvert_{x=4} = 4^2 + 1 = 17 \end{align}

Therefore, our lower bound $\alpha = 2$, and our upper bound $\beta = 17$. Therefore, we can carry on with evaluating our integral:

(3)

\begin{align} \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \int_{\alpha}^{\beta} \frac{1}{u} \: du \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \int_{2}^{17} u^{-1} \: du \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln \mid u \mid \biggr\rvert_{2}^{17} \\ \quad \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln 17 - \ln 2 \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln \frac{17}{2} \end{align}

Notice how we didn't need to make a substitution back for $u$ since our bounds of integration were already modified. Alternatively, we can skip the step of changing the limits of integration and proceed as follows:

(4)

\begin{align} \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \int_{\alpha}^{\beta} \frac{1}{u} \: du \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln \mid u \mid \biggr\rvert_{\alpha}^{\beta} \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln \mid x^2 + 1\mid \biggr\rvert_{1}^{4} \\ \int_{1}^{4} \frac{2x}{x^2 + 1} \: dx = \ln\frac{17}{2} \end{align}

As you can see, we have computed the same result regardless.

Note: Changing the lower/upper bounds or at least acknowledging that they're different is essential any time a substitution is made.

Example 1

Compute $\int_{1}^{2} 2x(x^2 - 1)^4 \: dx$.

First let $u = x^2 - 1$ so that $du = 2x \: dx$. Making the appropriate substitutions we get that:

(5)

\begin{align} \int_{1}^{2} 2x(x^2 - 1)^4 \: dx = \int_{\alpha}^{\beta} u^4 \: du \\ \end{align}

We calculate that $\alpha = u \: \rvert_{x = 1} = 0$ and $\beta = u \: \rvert_{x = 2} = 3$, and therefore:

(6)

\begin{align} \int_{1}^{2} 2x(x^2 - 1)^4 \: dx = \int_{0}^{3} u^4 \: du \\ \int_{1}^{2} 2x(x^2 - 1)^4 \: dx = \frac{u^5}{5} \biggr\rvert_{0}^{3} \\ \int_{1}^{2} 2x(x^2 - 1)^4 \: dx = \frac{243}{5} \end{align}