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Math Help - Frustrating integral

  1. #1
    Super Member angel.white's Avatar
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    Frustrating integral

    Well, I think I'm switching study subjects after this, because I've about had my fill.

    2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx

    I've tried about 10 different things, and can't seem to get anywhere.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Well, I think I'm switching study subjects after this, because I've about had my fill.

    2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx

    I've tried about 10 different things, and can't seem to get anywhere.
    a substitution of u = 1 - x should take care of this
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  3. #3
    Super Member angel.white's Avatar
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    Perhaps I'm doing it wrong:

    Substitution definitions:
    u=1-x

    \frac{du}{dx}=-1

    -du=dx

    Origional equation:
    2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx

    Substitute:
    2\pi \int_{0}^{1} (1+x) \sqrt{u} (-1)du

    I still have 1+x leftover. But I am pretty tired now, I've been looking at these all day, I'm probably overlooking something obvious
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Perhaps I'm doing it wrong:

    Substitution definitions:
    u=1-x

    \frac{du}{dx}=-1

    -du=dx

    Origional equation:
    2\pi \int_{0}^{1} (1+x) \sqrt{1-x} dx

    Substitute:
    2\pi \int_{0}^{1} (1+x) \sqrt{u} (-1)du

    I still have 1+x leftover. But I am pretty tired now, I've been looking at these all day, I'm probably overlooking something obvious
    so change the 1 + x into a function of u.

    u = 1 - x

    \Rightarrow x = 1 - u

    \Rightarrow x + 1 = 2 - u
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