Question

The value of the integral
${\int }_{-1}^1\{\frac{x^{2013}}{e^{\left|x\right|}(x^2+\cos x)}+\frac{1}{e^{\left|x\right|}}\}dx$ is equal to

• A. $0$
• B. $1-e^{-1}$
• C. $2e^{-1}$
• D. $2(1-e^{-1})$

Answer 1

The correct option is D $2(1-e^{-1})$

Let $I={\int }_{-1}^1\{\frac{x^{2013}}{e^{\left|x\right|}(x^2+\cos x)}+\frac{1}{e^{\left|x\right|}}\}dx$
$\Rightarrow$ $I={\int }_{-1}^1\frac{x^{2013}}{e^{\left|x\right|}(x^2+\cos x)}dx+{\int }_{-1}^1\frac{1}{e^{\left|x\right|}}dx$
Here $\frac{x^{2013}}{e^{\left|x\right|}\left(x^2\cos x\right)}$ is an odd function
and $\frac{1}{e^{\left|x\right|}}$ is even function
$\because {\int }_{-a}^{a}f\left(x\right)dx=\{\begin{array}{c}2{\int }_0^{a}f\left(x\right)dx;f\left(x\right)iseven\\ 0,f\left(x\right)isodd\end{array}$
$\therefore I=0+2{\int }_0^1{e}^{-x}dx=-2{\left(e^{-x}\right)}_0^1=-2\left(e^{-1}1\right)$
$=2(1-e^{-1})$