Question

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

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

Answer 1

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

Using the definite integral property;

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{-1}^1\{\frac{x^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx...eqn1$

Using the above mentioned property;

$I={\int }_{-1}^1\{\frac{{(-x)}^{2013}}{e^{|-x|}({(-x)}^2+\cos (-x))}+\frac{1}{e^{|-x|}}\}dxI={\int }_{-1}^1\{\frac{{-x}^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx...eqn2$

Adding $eqn1$ and $eqn2$;

$2I={\int }_{-1}^1\{\frac{x^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx+{\int }_{-1}^1\{\frac{{-x}^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx2I={\int }_{-1}^1\{\frac{x^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}+\frac{{-x}^{2013}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx2I={\int }_{-1}^1\left\{\frac{2}{e^{|x|}}\right\}dxI={\int }_{-1}^1\left\{\frac{1}{e^{|x|}}\right\}dx=2{\int }_0^1\left\{\frac{1}{e^x}\right\}dxI=2{\int }_0^1{e}^{-x}dx=-2{\left[e^{-x}\right]}_0^1=2[1-\frac{1}{e}]$.