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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 :

  1. (1) $-1$

  2. (2) 1

  3. (3) $-\frac{5}{2}$

  4. (4) $-\frac{1}{2}$


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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