Question

The value of the integral
$\underset{−1}{\overset{+1}{∫}}\left\{\frac{x^{2013}}{e^{âˆ\pounds xâˆ\pounds }\left(x^2+cosx\right)}+\frac{1}{e^{âˆ\pounds xâˆ\pounds }}\right\}dx$
is equal to

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

Answer 1

Answer: (C)

Let $I={∫}_{−1}^1\left\{\frac{x^{2013}}{e^{âˆ\pounds xâˆ\pounds }\left(x^2+\cos x\right)}+\frac{1}{e^{âˆ\pounds xâˆ\pounds }}\right\}dx$
$<br/>⇒I={∫}_{−1}^1\frac{x^{2013}}{e^{âˆ\pounds xâˆ\pounds }\left(x^2+\cos x\right)}dx+{∫}_{−1}^1\frac{1}{e^{âˆ\pounds xâˆ\pounds }}dx<br/>$
Here, $\frac{x^{2013}}{e^{âˆ\pounds xâˆ\pounds }\left(x^2+\cos x\right)}$ is an odd function
and $\frac{1}{âˆ\pounds xâˆ\pounds }$ is an even function.
$<br/>\{âˆ\mu {∫}_{−a}^{a}f(x)dx=\left\{\begin{array}{cc}<br/>2{∫}_0^{a}f(x)dx;& f(x)\text{ is even }\\ <br/>0,& f(x)\text{ is odd }<br/>\end{array}\right\}<br/>$
$<br/>∴l=0+2{∫}_0^1{e}^{−x}dx=−2{\left(e^{−x}\right)}_0^1=−2\left(e^{−1}1\right)<br/>$
$<br/>=2\left(1−e^{−1}\right)<br/>$