How do you find the integral of $t^{3} e^{- t^{2}}$ $t^3 e^(-t^2)$ ?

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Answer 1 · 2018-05-01T07:55:26

$\int t^{3} e^{- t^{2}} d t = - \frac{1}{2} e^{- t^{2}} (t^{2} + 1) + c$ $intt^3e^(-t^2) dt=-1/2e^(-t^2)(t^2+1)+c$ , where $c$ $c$ is a constant

Let $t^{2} = x$ $t^2=x$ , then $2 t d t = d x$ $2tdt=dx$ and

$\int t^{3} e^{- t^{2}} d t = \frac{1}{2} \int x e^{- x} d x$ $intt^3e^(-t^2) dt=1/2intxe^(-x)dx$

Now let us use integration by parts, which states that

$intu(x)v'(x)dx=u(x)v(x)-intv(x)u'(x)dx$

Let $u=x$ and $dv=e^-x$ , then $du=1$ and $v=-e^(-x)$

and hence $intxe^(-x)dx=-xe^(-x)+inte^(-x)dx$

= $-xe^(-x)-e^(-x)+c_1$ where $c_1$ is a constant.

Hence $intt^3e^(-t^2) dt=1/2intxe^(-x)dx$

= $1/2[-xe^(-x)-e^(-x)+c_1]$

= $-1/2xe^(-x)-1/2e^(-x)+c_1/2$

= $-1/2t^2e^(-t^2)-1/2e^(-t^2)+c$

= $-1/2e^(-t^2)(t^2+1)+c$ , where $c$ is another constant

How do you find the integral of t^3 e^(-t^2)t3e−t2t^3 e^(-t^2)?