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Math Help - Integration by Parts problem

  1. #1
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    Integration by Parts problem

    I've been struggling with integration by parts. Can anyone explain these problems to me?





    Help is very much appreciated. Thank you!
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  2. #2
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    Quote Originally Posted by Kimmy2 View Post
    I've been struggling with integration by parts. Can anyone explain these problems to me?





    Help is very much appreciated. Thank you!
    Recall that

    \int{u\,dv} = uv - \int{v\,du}.

    For Q.1 let u = \ln{t} so that du = \frac{1}{t}, and let dv = 7t^2 so that v = \frac{7}{3}t^3.


    Thus

    \int{7t^2 \ln{t}\,dt} = \frac{7}{3}t^3\ln{t} - \int{\frac{7}{3}t^3\,\frac{1}{t}\,dt}

     = \frac{7}{3}t^3\ln{t} - \frac{7}{3}\int{t^2\,dt}

     = \frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3 + C.


    Can you now find \int_1^e{7t^2\ln{t}\,dt}?
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  3. #3
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    For the second problem, set u = t and dv = e^{-t}dt
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  4. #4
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    So, I have:

    u=t
    du=dx

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

    And then:

    -te^(-t) - the integral from 6 to 0 of -e^(-t)dx

    And then if I take the integral of -e^-tdx, I get -ln(e^-t)?

    Is this okay, so far?
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  5. #5
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    Quote Originally Posted by Kimmy2 View Post
    So, I have:

    u=t
    du=dx

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

    And then:

    -te^(-t) - the integral from 6 to 0 of -e^(-t)dx

    And then if I take the integral of -e^-tdx, I get -ln(e^-t)?

    Is this okay, so far?
    A few issues.

    u = t so du = dt, not dx
    Also
    <br />
\int{-e^{-t}dt} = e^{-t}
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  6. #6
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    So, after I plug in all the u and v values, I'm getting:

    t(-e^-t) - the integral of -e^(-t)

    Once I integrate that last part, I'm getting:

    t(-e^-t) - e^(-t)

    Is this right?
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  7. #7
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    Yes that is correct. Just plug in the bounds now.
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  8. #8
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    I have plugged in the numbers multiple times, and my homework keeps telling me that I'm wrong.

    Some of the answers I got were:
    -0.1983001 and 1.01239376

    EDIT!
    I got it, thanks so much for your help!!
    Last edited by Kimmy2; November 9th 2009 at 07:52 PM.
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  9. #9
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    Quote Originally Posted by Kimmy2 View Post
    I've been struggling with integration by parts. Can anyone explain these problems to me?





    Help is very much appreciated. Thank you!
    Q.1

    \int_1^e{7t^2\ln{t}\,dt} = \left[\frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3\right]_1^e

     = \left[\frac{7}{3}e^3\ln{e} - \frac{7}{9}e^3\right] - \left[\frac{7}{3}(1)^3 \ln{1} - \frac{7}{9}(1)^3\right]

     = \left[\frac{7}{3}e^3 - \frac{7}{9}e^3\right] - \left[- \frac{7}{9}\right]

     = \frac{14}{9}e^3 + \frac{7}{9}.


    Q.2

    \int_0^6{t e^{-t}\,dt} = \left[-te^{-t} - e^{-t}\right]_0^6

    = \left[-6e^{-6} - e^{-6}\right] - \left[-0e^{-0} - e^{-0}\right]

     = -7e^{-6} - 1.
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