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

Question

Christopher Allers · Accepted Answer

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

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#

William Abdalla · Answer

undefined To evaluate the integral of ∫ t^2 * e^(-5t) dt, you can use integration by parts. The formula for integration by parts is:

∫ u dv = u * v - ∫ v du

In this case, let u = t^2 and dv = e^(-5t) dt. Then, differentiate u to get du and integrate dv to get v:

du = 2t dt
v = (-1/5) * e^(-5t)

Now, apply the integration by parts formula:

∫ t^2 * e^(-5t) dt = t^2 * (-1/5) * e^(-5t) - ∫ (-1/5) * e^(-5t) * 2t dt

= (-t^2/5) * e^(-5t) + (2/5) * ∫ t * e^(-5t) dt

Now, you can use integration by parts again for the remaining integral:

Let u = t and dv = e^(-5t) dt. Then, du = dt and v = (-1/5) * e^(-5t).

Apply the integration by parts formula:

∫ t * e^(-5t) dt = t * (-1/5) * e^(-5t) - ∫ (-1/5) * e^(-5t) * dt

= (-t/5) * e^(-5t) + (1/5) * ∫ e^(-5t) dt

= (-t/5) * e^(-5t) - (1/25) * e^(-5t) + C

Substitute this back into the previous expression:

(-t^2/5) * e^(-5t) + (2/5) * [(-t/5) * e^(-5t) - (1/25) * e^(-5t)] + C

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

So, the integral of ∫ t^2 * e^(-5t) dt is (-t^2/5) * e^(-5t) - (2/25) * t * e^(-5t) + (2/125) * e^(-5t) + C, where C is the constant of integration.