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 $x^{2} =t \; 2xdx =dt$
$\Rightarrow \frac{1}{2} \int t^{2} .e^{-t}dt$
$= \frac{1}{2} \left[-t^{2} .e^{-t} + \int2t.e^{-t}.dt\right]$
$= \frac{1}{2}\left(-t^{2} .e^{-t}\right) + \left(-t.e^{-t} +\int1.e^{-t}.dt\right) = - \frac{t^{2}e^{-t}}{2} - te^{-t} -e^{-t}$
$= \left(- \frac{t^{2}}{2} -t-1\right)e^{-t} = \left(- \frac{x^{4}}{2} - x^{2}1\right)e^{-x^2}+C$
$g\left(x\right) = -1 -x^{2} - \frac{x^{4}}{2} +ke^{x^2}$
for $k = 0$
$g\left(-1\right) =- 1 -1 -\frac{1}{2} = - \frac{5}{2}$