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

1 Answer
Dec 18, 2016

-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