Q. 53 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_{-1}^1\frac{e^x-x{e}^x}{e^{2x}}dx$

Step-by-Step Solution

Answer

${\int }_{-1}^1\frac{e^x-x{e}^x}{e^{2x}}dx=\frac{1}{e}+e$.

Step 1. Given information.

A definite integral is given as ${\int }_{-1}^1\frac{e^x-x{e}^x}{e^{2x}}dx$.

Step 2. Using the Fundamental theorem of Calculus.

Let

$$\begin{gathered} f\left(x\right)=x \\ g\left(x\right)=e^x \end{gathered}$$

then

$$\begin{gathered} \frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)} \\ =\frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2} \\ =\frac{e^x-x{e}^x}{e^{2x}} \end{gathered}$$

Now we get

$$\begin{gathered} {\int }_{-1}^1\frac{e^x-x{e}^x}{e^{2x}}dx \\ ={\left[\frac{x}{e^x}\right]}_{-1}^1 =[\frac{1}{e}-\frac{-1}{e^{-1}}] \\ =\frac{1}{e}+e \end{gathered}$$

The exact value of the given definite integral is $\frac{1}{e}+e$.

Step 3. The graph to verify the answer is

The solution is area under graph which is

$$\begin{gathered} a\approx 3.086161 \\ \approx e+\frac{1}{e} \end{gathered}$$