Question

How do you evaluate the integral of `int t^2*e^(-5*t) dt`?

Answer 1

`-t^2/5e^(-5t)-(2t)/25e^(-5t)-2/125e^(-5t)+C`

Explanation:

We will use integration by parts. This takes the form `intudv=uv-intvdu`. When we choose values for `u` and `dv`, we want our value of `u` to become simpler as we differentiate it.

So, let `u=t^2` and `dv=e^(-5t)dt`. So, `du=2tdt` and `v=inte^(-5t)dt=-1/5e^(-5t)`.

So:

`intt^2e^(-5t)dt=uv-intvdu`

`color(white)(intt^2e^(-5t)dt)=t^2(-1/5e^(-5t))-int(-1/5e^(-5t))(2tdt)`

`color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)+int2/5te^(-5t)dt`

Again, we need integration by parts. Let `u=2/5t` so `du=2/5dt`. Again let `dv=e^(-5t)dt` so again `v=-1/5e^(-5t)`.

`color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)+(uv-intvdu)`

`color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)+2/5t(-1/5e^(-5t))-int(-1/5e^(-5t))(2/5dt)`

`color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)-(2t)/25e^(-5t)+int2/25e^(-5t)dt`

Using the integral from before, `inte^(-5t)dt=-1/5e^(-5t)`:

`color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)-(2t)/25e^(-5t)-2/125e^(-5t)+C`