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Math Help - Easy Integral

  1. #1
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    Easy Integral

    I have an integral that I think is probably super easy, but I'm just totally missing the point...

    integrate: x(1-x)^10

    I can't substitute anything here, can I? It looks like questions I've done a thousand times before, but this one just isn't clicking...
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  2. #2
    o_O
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    Sure you can: {\color{red}u = 1 - x} \ \Rightarrow \ du = - dx \ \Leftrightarrow \ {\color{blue}-du = dx}

    Also note that if you rearrange the red, we have: \color{magenta}x = 1 - u

    So: \int {\color{magenta}x}({\color{red}1-x})^{10} \ {\color{blue}dx} = \int ({\color{magenta}1-u}) {\color{red}u}^{10} ({\color{blue}-du}) = \cdots
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    Quote Originally Posted by o_O View Post
    Sure you can: {\color{red}u = 1 - x} \ \Rightarrow \ du = - dx \ \Leftrightarrow \ {\color{blue}-du = dx}

    Also note that if you rearrange the red, we have: \color{magenta}x = 1 - u

    So: \int {\color{magenta}x}({\color{red}1-x})^{10} \ {\color{blue}dx} = \int ({\color{magenta}1-u}) {\color{red}u}^{10} ({\color{blue}-du}) = \cdots

    Ok, that is tricky. So then I would multiply ({\color{magenta}1-u}) {\color{red}u}^{10}, and then integrate?
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  4. #4
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    Yep: (1-u)u^{10} = u^{10} - u^{11}

    And I'm sure you're familiar with integrating powers
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  5. #5
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    Quote Originally Posted by Hellreaver View Post
    Ok, that is tricky. So then I would multiply ({\color{magenta}1-u}) {\color{red}u}^{10}, and then integrate?
    You can also do it by parts. Although, that's probably more time consuming.
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