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Math Help - Area of shaded region

  1. #1
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    Area of shaded region

    I posted a question similar to this a minute ago, but I wanted to double check and make sure I'm doing it right. The question is find the shaded region of 7x(x^2-64) There is a picture, but I can't display it here. The bounds are 0 to 8 and the second equation is y = 0. So, you take the top curve (0) - the bottom curve 7x(x^2-64). Leaving you with -7x^3-448x. From there you integrate which gives -7/4x^4 - 224x^2. This is where I get confused with definite integration. Do I put in 8 for the first x and then 0 for the second x? E.g.  -7/4(8)^4 - 224(0)^2 Or do I do two whole separate equations E.g. (-7/4(8)^4 - 224(8)^2) - (-7/4(0)^4 - 224(0)^2) . Can someone explain? Thanks
    Last edited by zsig013; May 7th 2008 at 04:57 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zsig013 View Post
    I posted a question similar to this a minute ago, but I wanted to double check and make sure I'm doing it right. The question is find the shaded region of 7x(x^2-64) There is a picture, but I can't display it here. The bounds are 0 to 8 and the second equation is y = 0. So, you take the top curve (0) - the bottom curve 7x(x^2-64). Leaving you with -7x^3-448x. From there you integrate which gives -7/4x^4 - 224x^2. This is where I get confused with definite integration. Do I put in 8 for the first x and then 0 for the second x? E.g.  -7/4(8)^4 - 224(0)^2 Or do I do two whole separate equations E.g. (-7/4(8)^4 - 224(8)^2) - (-7/4(0)^4 - 224(0)^2) . Can someone explain? Thanks
    Remember to evaluate the definite integral, we need to use the Fundamental Theorem of Calculus:
    \int_{a}^{b}f(x)\,dx=F(b)-F(a)
    where F(x) is the antiderivative of f(x).

    In your case, we have the antiderivative -\frac{7}{4}x^4-224x^2. Then we need to find F(8)-F(0)=(-\frac{7}{4}(8)^4-224(8)^2)-(-\frac{7}{4}(0)^4-224(0)^2).

    I hope this clarified things!!!
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