Question

Consider the integral ${\int }_{-2}^2{e}^{5-3x}dx$.

(a) Solve this integral by using u-substitution while keeping the limits of integration in terms of x.

(b) Solve the integral again with u-substitution, this time changing the limits of integration to be in terms of u.

Short answer

(a) The value of integral by using u-substitution while keeping the limits of integration in terms of x is ${\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}\left[e^{-1}-e^{11}\right]$.

(b) The value of integral again with u-substitution, this time changing the limits of integration to be in terms of u is ${\int }_{-2}^2{e}^{5-3x}\mathrm{d}x=-\frac{1}{3}\left[e^{-1}-e^{11}\right]$.

Step-by-step solution

Step 1. Given Information 

Consider the integral ${\int }_{-2}^2{e}^{5-3x}dx$.

(a) Solve this integral by using u-substitution while keeping the limits of integration in terms of x.

(b) Solve the integral again with u-substitution, this time changing the limits of integration to be in terms of u.

Part (a) Step 1. Now solving this integral by using u-substitution while keeping the limits of integration in terms of x. 

Let

$$\begin{gathered} u=5-3x \\ \frac{du}{dx}=-3 \\ -\frac{1}{3}du=dx \end{gathered}$$

Part (a) Step 2. This substitution changes the integral into 

$$\begin{gathered} {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}{\int }_{-2}^2{e}^u\mathrm{d}u \\ {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}{\left[e^u\right]}_{-2}^2 \\ {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}{\left[e^{5-3x}\right]}_{-2}^2 \\ {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}\left[e^{5-3\times 2}-e^{5-3\times (-2)}\right] \\ {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}\left[e^{5-6}-e^{5+6}\right] \\ {\int }_{-2}^2{e}^{5-3x}dx=-\frac{1}{3}\left[e^{-1}-e^{11}\right] \end{gathered}$$

Part (b) Step 2. Now solve the integral again with u-substitution, this time changing the limits of integration to be in terms of u.

Let

$$\begin{gathered} u=5-3x \\ \frac{du}{dx}=-3 \\ -\frac{1}{3}du=dx \end{gathered}$$

Part (b) Step 2. We will now write the limits of integration (x=-2 and x=2) in terms of the new variable u.

When $x=-2$ we have

$$\begin{gathered} u=5-3x \\ u(-2)=5-3(-2) \\ u(-2)=5+6 \\ u(-2)=11 \end{gathered}$$

When $x=2$we have

$$\begin{gathered} u=5-3x \\ u\left(2\right)=5-3\left(2\right) \\ u\left(2\right)=5-6 \\ u\left(2\right)=-1 \end{gathered}$$

Part (a) Step 3. This substitution changes the integral into 

$$\begin{gathered} {\int }_{-2}^2{e}^{5-3x}\mathrm{d}x=-\frac{1}{3}{\int }_{11}^{-1}{e}^u\mathrm{d}u \\ {\int }_{-2}^2{e}^{5-3x}\mathrm{d}x=-\frac{1}{3}{\left[e^u\right]}_{11}^{-1} \\ {\int }_{-2}^2{e}^{5-3x}\mathrm{d}x=-\frac{1}{3}\left[e^{-1}-e^{11}\right] \end{gathered}$$