Question

$\underset{-1}{\overset{1}{\int }}{e}^{\left|x\right|}dx=$ ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{-1}{\overset{1}{\int }}{e}^{\left|x\right|}dx \\ \mathrm{Let}f\left(x\right)=e^{\left|x\right|} \\ f\left(-\mathrm{x}\right)=e^{\left|-x\right|} \\ =e^{\left|x\right|} \\ =f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,\mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=2\underset{0}{\overset{1}{\int }}{e}^{\left|x\right|}dx \\ =2\underset{0}{\overset{1}{\int }}{e}^{x}dx \\ =2{\left(e^x\right)}_0^1 \\ =2\left(e^1-e^0\right) \\ =2\left(e-1\right) \end{gathered}$$​


​Hence, $\underset{-1}{\overset{1}{\int }}{e}^{\left|x\right|}dx=\underline{2\left(e-1\right)}.$