If [ x^5 e^ - x^2dx = g( x )e^ - x^2 + c], where c is a constant of integration, then [g(-…

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 :

Accepted Answer

Let x2 = t

$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$

$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$

$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$

$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x^2} - 1} \right){e^{ - {x^2}}} + c$$

Then $$g(x) =  - {{{x^4}} \over 2} - {x^2} - 1$$

$$ \Rightarrow g( - 1) =  - {1 \over 2} - 1 - 1$$

$$ \Rightarrow  - {5 \over 2}$$

Solutions · 1