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Thread: 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 $\displaystyle a(t) = \frac{t}{10}e^{\frac{t}{10}}$ ? v(t) would be the integral is v and a were velocity and acceleration respectively.

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

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

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

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

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

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

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

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

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

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


    I'll do it in conventional terms . . .


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


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


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

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