Question

If $\int x^5{e}^{-x^2}dx=g\left(x\right)e^{-x^2}+c,$ where $c$ is a constant of integration, then $g(-1)$ is equal to :

• A. $-\frac{5}{2}$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A $-\frac{5}{2}$

Given,
$I=\int x^5{e}^{-x^2}dx$
Let $x^2=t$
$=\frac{1}{2}\int t^2{e}^{-t}dt$
$=\frac{1}{2}[-t^2{e}^{-t}+\int 2t{e}^{-t}dt]$
$=\frac{-t^2{e}^{-t}}{2}-t{e}^{-t}-e^{-t}$
$=(-\frac{x^4}{2}-x^2-1)e^{-x^2}+c...\left(1\right)$
Now compairing with the equation $g\left(x\right)e^{-x^2}+c$
$g\left(x\right)=-\frac{1}{2}(x^4+2x^2+2)$
So, $g(-1)=-\frac{1}{2}(1+2+2)=-\frac{5}{2}$