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Math Help - integral calculus

  1. #1
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    integral calculus

    find the area bounded by the curve y^2=2x+3 and y=x
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  2. #2
    Eater of Worlds
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    Here's the graph. You should be able to find your limits from it.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  3. #3
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    Hello, alderon!

    Find the area bounded by the curves: . \begin{array}{cccc}y^2 & = & 2x+3 & {\color{blue}[1]} \\ y & = & x & {\color{blue}[2]}\end{array}
    We'll do this one "sideways".

    Equation {\color{blue}[1]} is: . x \;=\;\frac{1}{2}y^2 - \frac{3}{2}

    Code:
                     |     /
                     |    *
                     |*::/
                  *::|::/
               *:::::|:/
             *:::::::|/
        - - *:-:-:-:-*- - - -
             *::::::/|
               *:::/ |
                  *  |
                 /   |*
                     |    *

    Substitute {\color{blue}[2]} into {\color{blue}[1]}: . y^2 \;=\;2y + 3\quad\Rightarrow\quad y^2 - 2y - 3 \;=\;0

    . . which factors: . (y + 1)(y - 3) \;=\;0

    . . and has roots: . y \:=\:\text{-}1,\,3


    The area is: . A \;=\;\int_{-1}^3\left[y - \left(\frac{1}{2}y^2 - \frac{3}{2}\right)\right]\,dy

    Got it?

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  4. #4
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    i got it

    thanks... now the solution is complete
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