Question

The value of ${\int }_{-2}^2\left[1+2\sin \left(x\right)\right]e^{\left|x\right|}dx$ is

• A. $0$
• B. $e^2-1$
• C. $2\left(e^2-1\right)$
• D. $1$

Answer 1

The correct option is C

$2\left(e^2-1\right)$

Explanation for the correct option.

Compute the required value.

Given: ${\int }_{-2}^2\left[1+2\sin \left(x\right)\right]e^{\left|x\right|}dx$

$\Rightarrow {\int }_{-2}^2{e}^{\left|x\right|}dx+{\int }_{-2}^2\left[1+2\sin \left(x\right)\right]dx$

$e^{\left|x\right|}=\left\{\begin{array}{l}{e}^x,0\le x<2\\ e^{-x},-2\le x<0\end{array}\right.$

$\Rightarrow {\int }_{-2}^0{e}^{-x}dx+{\int }_0^2{e}^{x}dx+{\int }_{-2}^2\left[e^{x}2\sin \left(x\right)\right]dx$

${\int }_{-2}^2\left[1+2\sin \left(x\right)\right]dx$ is an even function and ${\int }_{-2}^2\left[e^{x}2\sin \left(x\right)\right]dx$ will be equal to $0$.

$$\begin{gathered} \Rightarrow -{\left|e^{-x}\right|}_{-2}^0+{\left|e^x\right|}_0^2+0 \\ \Rightarrow 1{-}^0+e^{-(-2)}+e^2-e^0 \\ \Rightarrow -1+e^2+e^2-1 \\ \Rightarrow 2(e^2-1) \end{gathered}$$

Hence, option $C$ is the correct answer.