Question

If $\int x^5{e}^{-x^2}=g\left(x\right)e^{-x^2}+c$, 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)

$-\frac{5}{2}$

Let $x^2=t$

$2xdx=dt$

$\Rightarrow \frac{1}{2}\int t^2{e}^{-t}dt=\frac{1}{2}[\int {-t}^2{e}^{-t}+\int {2t}^{}{e}^{-t}]=\frac{1}{2}({-t}^2.e^{-t})+({-t}^{}.e^{-t}+\int 1.e^{-t}dt)=-\frac{{-t}^2.e^{-t}}{2}-{-t}^{}.e^{-t}-e^{-t}$

$=(-\frac{{-t}^2}{2}-t-1)e^{-t}=(-\frac{x^4}{2}-x^2-1)e^{-x^2}+C$

$g\left(x\right)=-1-x^2-\frac{x^4}{2}$

for $k=0$

$g(-1)=-1-1-\frac{1}{2}=-\frac{5}{2}$