Question

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

Answer 1

`intt^3e^(-t^2) dt=-1/2e^(-t^2)(t^2+1)+c`, where `c` is a constant

Explanation:

Let `t^2=x`, then `2tdt=dx` and

`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