Question

If $\int x^5{e}^{-x^2}dx=g\left(x\right)\cdot e^{-x^2}+C$ then the value of $g(-1)$ is?

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $-\frac{5}{2}$
• D. $\frac{e}{2}$

Answer 1

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

Put $x^2=t$
$2xdx=dt$
$\int t^2{e}^{-t}\frac{dt}{2}$
$=\frac{1}{2}[-t^2\cdot e^{-t}+2\int t{e}^{-1}dt]+c$
$=\frac{1}{2}[-t^2\cdot e^{-t}-2t{e}^{-t}+\int 2e^{-t}dt]+c$
$=\frac{1}{2}(-t^2{e}^{-t}-2(t{e}^{-1}+e^{-t}\left)\right)+c$
$=\frac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$
$g\left(x\right)=\frac{-(x^4+2x^2+2)}{2}$
$g(-1)=-\frac{5}{2}$.