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Math Help - integration question (again)

  1. #1
    Junior Member cinder's Avatar
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    integration question (again)

    Sorry, but it's been a while since I've done calculus.

    Evaluate the integral:

    \int re^{r/2}dr

    Do you let u = r and dv = e^{r/2}dr?

    From there, I'm not sure what to do.
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  2. #2
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     \int udv = uv - \int vdu

    Follow the formula.
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  3. #3
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    Hello, cinder!

    Evaluate the integral: . \int re^{\frac{r}{2}}dr

    Do you let: u = r and dv = e^{\frac{r}{2}}dr? . . . . Yes!

    Integration by parts . . .

    . . \begin{array}{ccc}u = r & dv = e^{\frac{r}{2}}dr \\ du = dr & v = 2e^{\frac{r}{2}} \end{array}

    And we have: . 2re^{\frac{r}{2}} - 2\int e^{\frac{r}{2}}\,dr \;=\;2re^{\frac{r}{2}} - 4e^{\frac{r}{2}} + C \;=\;2e^{\frac{r}{2}}(r - 2) + C

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  4. #4
    Junior Member cinder's Avatar
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    I really don't remember (or get) what you do to the u and dv to change them like that.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by cinder View Post
    Sorry, but it's been a while since I've done calculus.

    Evaluate the integral:

    \int re^{r/2}dr

    Do you let u = r and dv = e^{r/2}dr?

    From there, I'm not sure what to do.
    set u=r, dv=e^{r/2}, then:

    \int re^{r/2}dr=uv-\int v\,du=r\,2e^{r/2}-2\int e^{r/2}dr=r\,2e^{r/2}-4 e^{r/2}=2e^{r/2}(r-2)

    RonL
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