Q. 64 - 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.

${\int }_{-1}^1\left|e^{2x}-1\right|dx$

Step-by-Step Solution

Answer

The value of integral is $-\frac{e^2}{2}+1+\frac{1+2e^2}{2e^2}$ and the plot is

Step 1. Given information

An expression is given as ${\int }_{-1}^1\left|e^{2x}-1\right|dx$

Step 2. Evaluating integral

Evaluate the integral as,

$$\begin{gathered} {\int }_{-}^1\left|e^{2x}-1\right|dx \\ ={\int }_{-1}^0-e^{2x}+1dx+{\int }_0^1{e}^{2x}-1dx \\ ={\left[\frac{-e^{2x}}{2}\right]}_{-1}^0+[x{]}_{-1}^0+{\left[\frac{e^{2x}}{2}\right]}_0^1-[x{]}_0^1 \\ =\left[\frac{-e^{2\left(0\right)}}{2}+\frac{e^{2(-1)}}{2}\right]+[0+1]-\left[\frac{e^2}{2}-\frac{e^{2\left(0\right)}}{2}\right]-[1+0] \\ =\left[-\frac{1}{2}+\frac{e^{-2}}{2}+1\right]+\left[-\frac{e^2}{2}+\frac{1}{2}+1\right] \\ =\frac{1}{2}+\frac{1}{2e^2}+\frac{1}{2}-\frac{e^2}{2} \\ =-\frac{e^2}{2}+1+\frac{1+2e^2}{2e^2} \end{gathered}$$

And the plot is