Question:
If $\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+c$, where $c$ is a constant of integration, then $g(-1)$ is equal to :
Correct Option: , 3
Solution:
Let, $1=\int x^{2} \cdot e^{-x^{2}} d x$
Put $-x^{2}=t \Rightarrow-2 x d x=d t$
$1=\int \frac{t^{2} \cdot e^{t} d t}{(-2)}=\frac{-1}{2} e^{t}\left(t^{2}-2 t+2\right) c$
$\therefore g(x)=\frac{-1}{2}\left(x^{4}+2 x^{2}+2\right) \Rightarrow g(-1)=\frac{-5}{2}$
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