Question

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

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

Answer 1

Answer: (C)

Let $I=\int_{-1}^{1}\left\{\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x$
$
\Rightarrow I=\int_{-1}^{1} \frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x
$
Here, $\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}$ is an odd function
and $\frac{1}{|x|}$ is an even function.
$
\left\{\because \int_{-a}^{a} f(x) d x=\left\{\begin{array}{cc}
2 \int_{0}^{a} f(x) d x ; & f(x) \text { is even } \\
0, & f(x) \text { is odd }
\end{array}\right\}\right.
$
$
\therefore l=0+2 \int_{0}^{1} e^{-x} d x=-2\left(e^{-x}\right)_{0}^{1}=-2\left(e^{-1} 1\right)
$
$
=2\left(1-e^{-1}\right)
$