Mathematics

# If $\displaystyle\int x^5e^{-x^2}dx=g(x)\cdot e^{-x^2}+C$ then the value of $g(-1)$ is?

$-\dfrac{5}{2}$

##### SOLUTION
Put $x^2=t$
$2xdx=dt$
$\displaystyle\int t^2e^{-t}\dfrac{dt}{2}$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}+2\displaystyle\int te^{-1}dt]+c$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}-2te^{-t}+\displaystyle\int 2e^{-t}dt]+c$
$=\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$
$=\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$
$g(x)=\dfrac{-(x^4+2x^2+2)}{2}$
$g(-1)=-\dfrac{5}{2}$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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