How to solve an integration problem using integration by parts

[rgitelman]

A particle that moves along a straight line has velocity v(t) = t2e-t meters per second after t seconds. How far will it travel during the first t seconds.

This problem is from “calculus” by James Stewart 5e pg 517 # 61

I started the problem by using the formula for integration by parts. This is what I have.

∫udv = uv - ∫ v du

∫ t2e-t dt =

u = t2 dv = e-t
du = 2t v = -e-t

I don’t know where to go on from here. Can someone please show me the solution to this problem?

 

[Kummer]

\int t^2 e^{-t} dt

u=t^2 \mbox{ and }v'=e^{-t}\implies u'=2t \mbox{ and }v=-e^{-t}

\implies uv - \int u'vdt = -t^2e^{-t}+2\int te^{-t} dt

Continue ...

-Wolfgang.

 

[Steven60]

If u = t^2 then du is not 2t and dv is not e^-t

 

[cristo]

If u = t^2 then du is not 2t and dv is not e^-t

Why not? (Well, dv=-e^{-t} as shown above).

 

[rgitelman]

This is what I have:
∫t ^2 e ^-t dt
u = t ^2 dv = e ^-t
du = 2t v = -e ^-t

-t ^2 e ^-t + 2∫te ^-t dt
u = t dv = e ^-t
du = 1 dt v = -e ^-t

-t ^2 e ^-t + 2((t)(-e ^-t) - ∫(-e ^-t)(1)
-t ^2 e ^-t + 2(-te ^ -t) – (e^-t)
-t ^2 e ^-t – 2te^-t – 2e^-t
-e^-t (t^2 + 2t + 2)

The answer in the book says 2-e^-t (t^2 + 2t + 2)

I don't know what I'm doing wrong.

 

[janhaa]

integration

Your integration by parts is right, but the particle have moved the distance, d:

d=[-e^{-t}(t^2+2t+2)]_0^t=(-e^{-t}(t^2+2t+2)\,+\,e^0\cdot 2)=2\,-\,e^{-t}(t^2+2t+2)

 

[rgitelman]

thank you, i figured it out. c = 2