How do you evaluate the integral of int t^2*e^(-5*t) dt?
1 Answer
Explanation:
We will use integration by parts. This takes the form
So, let
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
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,
color(white)(intt^2e^(-5t)dt)=-t^2/5e^(-5t)-(2t)/25e^(-5t)-2/125e^(-5t)+C