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

Answer: (A)

Let $x^2=t2xdx=dt$
$\Rightarrow \frac{1}{2}\int t^2.e^{-t}dt$
$=\frac{1}{2}\left[-t^2.e^{-t}+\int 2t.e^{-t}.dt\right]$
$=\frac{1}{2}\left(-t^2.e^{-t}\right)+\left(-t.e^{-t}+\int 1.e^{-t}.dt\right)=-\frac{t^2{e}^{-t}}{2}-t{e}^{-t}-e^{-t}$
$=\left(-\frac{t^2}{2}-t-1\right)e^{-t}=\left(-\frac{x^4}{2}-x^{2}1\right)e^{-x^2}+C$
$g\left(x\right)=-1-x^2-\frac{x^4}{2}+k{e}^{x^2}$
for $k=0$
$g\left(-1\right)=-1-1-\frac{1}{2}=-\frac{5}{2}$