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Math Help - simple integration by parts

  1. #1
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    simple integration by parts

    How do you integrate (re^r/2)dr ?

    Here is what I have so far, I think I am making a mistake with the integration of e^r/2.

    (re^r/2)dr u = r dV = e^r/2dr
    du = dr V = (1/2)e^r/2

    uV - (integral of) Vdu

    so:

    (1/2)re^r/2 - (integral of)(1/2)e^r/2 dr

    (1/2)re^r/2 - (1/4)e^r/2 + c
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  2. #2
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    Quote Originally Posted by mech.engineer.major View Post
    How do you integrate (re^r/2)dr ?

    Here is what I have so far, I think I am making a mistake with the integration of e^r/2.

    (re^r/2)dr u = r dV = e^r/2dr
    du = dr V = (1/2)e^r/2

    uV - (integral of) Vdu

    so:

    (1/2)re^r/2 - (integral of)(1/2)e^r/2 dr

    (1/2)re^r/2 - (1/4)e^r/2 + c
    Is the question \int{re^{\frac{r}{2}}\,dr} or \int{\frac{re^r}{2}\,dr}?
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    the first of yours
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    To find \int{re^{\frac{r}{2}}\,dr}

    Remember that

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


    Let u = r so that \frac{du}{dr} = 1.

    Let \frac{dv}{dr} = e^{\frac{r}{2}} so that v = 2e^{\frac{r}{2}}.


    Therefore

    \int{re^{\frac{r}{2}}\,dr} = 2re^{\frac{r}{2}} - \int{2e^{\frac{r}{2}}\,dr}

     = 2re^{\frac{r}{2}} - 4e^{\frac{r}{2}} + C.
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  5. #5
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    yeah thats what I thought I did wrong. I needed to divide e^(r/2) by the derivative of the exponent to get 2e^(r/2) instead of multiplying to get (1/2)e^(r/2).

    thanks for your help
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