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    Attachment 613528The question is at the top and the substitution on the top right... what have I done wrong here, because I used the definite integral function on my calculator to find out what the right answer is (0.1523809524).
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    RDKGames help please
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    Anyone?
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    Let u=x+1. Then it follows x=u-1, and \frac{du}{dx} = 1, hence du=dx. When x=0 \implies u=1 and when x=-1 \implies u=0.

    Now let I = \int_{0}^{-1}x^2\sqrt{x+1}dx. By using our above principles, it follows that I can be written as I=\int_{0}^{1}(u-1)^2u^{1/2}du. Expanding the parentheses gives I=\int_{0}^{1}(u^2-2u+1)u^{1/2}du, hence by further expansion we get I=\int_{0}^{1}u^{5/2}-2u^{3/2}+u^{1/2}du. I'm sure you can figure out the rest.
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    you integrated when doing du/dx
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    (Original post by potassiumnitrate)
    you integrated when doing du/dx
    Omg yeah.... I'm so stupid thanks
 
 
 
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