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Help with integration problem

  1. Jul 25, 2007 #1
    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?
     
  2. jcsd
  3. Jul 25, 2007 #2
    [tex]\int t^2 e^{-t} dt[/tex]

    [tex]u=t^2 \mbox{ and }v'=e^{-t}\implies u'=2t \mbox{ and }v=-e^{-t}[/tex]

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

    Continue ....

    -Wolfgang.
     
  4. Jul 25, 2007 #3
    If u = t^2 then du is not 2t and dv is not e^-t
     
  5. Jul 25, 2007 #4

    cristo

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    Staff Emeritus
    Science Advisor

    Why not? (Well, dv=-e^{-t} as shown above).
     
  6. Jul 25, 2007 #5
    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.
     
  7. Jul 25, 2007 #6
    integration

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

    [tex]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)[/tex]
     
  8. Jul 25, 2007 #7
    thank you, i figured it out. c = 2
     
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