Use integration by parts to evaluate the definite integral. t*e^-tdt from 0,4

should I use u=t
dv=e^-tdt

how would I proceed? really not understanding this. :(

yes, go as you have it. u=t and dv = e^-t dt

using the formula for integration by parts-- you should get:

-4e^(-4) - int (-e^(-t) dt), 0,4 and continuing you get

(-5/e^4) + 1

Yes. Use u=t and dv=e^-tdt. You need to take the derivative of u (which is dt) and integrate dv. Remember the formula is as follows: ∫udv=uv-∫vdu. (I had to memorize that for my calculus class.

∫e^(-t)dt Let w=-t, -dw=dt
-∫e^(u)du=-e^u=-e^(-t) =>v

∫te^(-t)dt=-te^(-t)-∫e^(-t)dt
4
-te^(-t)+e^(-t)| There is no Calculus after this step. Everything else is Algebra.
0
[-(4)e^(-(4))+e^(-(4))]-[-(0)e^(-(0))+…
[-3e^(-4)]-[-0(1)+1]
[-3e^4]-[0+1]
[-3e^4]-[1]
-3e^4-1

Approximately 99% of Calculus is Algebra. Keep that in mind.

∫t e^(-t) dt from 0 to 4

u = t
du = dt

dv = e^(-t)
v = - e^(-t)

∫ u dv = uv - ∫v du

=> ∫t e^(-t) = -t e^(-t) - ∫- e^(-t) dt

=> ∫t e^(-t) = -t e^(-t) + ∫ e^(-t) dt

=> ∫t e^(-t) = -t e^(-t) - e^(-t) from 0 to 4

= -e^(-t)( t + 1) from 0 to 4

= -e^(-4)(4 + 1) + e^(0) (0 + 1)

= 1 - 5 e^(-4)

= 0.981

should I use u=t
du=dt
dv=e^-tdt
v=-e^-t

yes:

∫ u dv = uv - ∫ vdu
∫ t*e^-tdt = -t e^-t - ∫ [-e^-t] dt
= -t e^-t + ∫ e^-t dt
= -t e^-t -e^-t + C