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Math Help - Integration by parts

  1. #1
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    Integration by parts

    I am struggling with this question
    a(t)= t/10e^t/10 so v(t) I think is the integral of this?
    can somebody explain this to me using integration by parts
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  2. #2
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    Quote Originally Posted by Jaffa View Post
    I am struggling with this question
    a(t)= t/10e^t/10 so v(t) I think is the integral of this?
    can somebody explain this to me using integration by parts
    Do you mean a(t) = \frac{t}{10}e^{\frac{t}{10}} ? v(t) would be the integral is v and a were velocity and acceleration respectively.

    I would substitute b = \frac{t}{10} but it doesn't really matter.

    therefore \frac{db}{dt} = 0.1 and dt = 10db

    a(t) = be^b hence v(t) = \int{be^b \cdot 10db}

    Using integration by parts letting u = b and \frac{dv}{db} = e^b so that

    \frac{du}{db} = 1 and v = e^b

    I = uv - \int{v\frac{du}{db}} = be^b - \int{e^b} = e^b(b-1)

    remembering that b = \frac{t}{10} we get

    e^{\frac{t}{10}}(1-\frac{t}{10}) + C
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  3. #3
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    Hello, Jaffa!

    \int \frac{t}{10}e^{\frac{t}{10}}dt

    We have: . \frac{1}{10}\int t\, e^{\frac{t}{10}} dt


    I'll do it in conventional terms . . .


    . . \begin{array}{ccccccc}u &=&t & & dv &=& e^{\frac{t}{10}}dx \\ du &=& dx & & v &=& 10e^{\frac{t}{10}} \end{array}


    Then: . \frac{1}{10}\bigg[t\cdot10e^{\frac{t}{10}} - \int10e^{\frac{t}{10}}dt\bigg] + C \;=\;\frac{1}{10}\bigg[10te^{\frac{t}{10}} - 100e^{\frac{t}{10}}\bigg] + C


    . . . = \;e^{\frac{t}{10}}(t-10) + C

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