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Thread: Area between 2 curves (well...3)

  1. #1
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    Area between 2 curves (well...3)

    $\displaystyle y = 4x^2$
    $\displaystyle y = 8x^2$
    $\displaystyle 2x+y=6$

    x > or equal to 0

    I have found the intersection points of 4x^2 and 2x+y=6 to be at (-1.5,9) and (1,4) and the intersection points of 8x^2 and 2x+y=6 to be at (-1,4) and (.75,4.5)

    I set up my integral to be from -1.5 to 1 $\displaystyle 8x^2-4x^2$ = 35/6

    I also tried setting up my integral from 0 to 1 $\displaystyle 8x^2-4x^2$ = 4/3

    But...this is wrong...can anyone help plz? thx
    Last edited by silencecloak; Feb 2nd 2009 at 05:59 AM.
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  2. #2
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    Hello, silencecloak!

    I think you need a better sketch . . .


    Find the area of the region enclosed by: . $\displaystyle \begin{array}{c}y \:= \:4x^2 \\y \:= \:8x^2 \\y\:=\:\text{-}2x+6 \\ x \:\geq\:0 \end{array}$

    The area is in quadrant 1, between two parabolas and a slanted line.
    Code:
            |
            |       \   *  y = 8x
            |        \
            |         \             *  y = 4x
            |          *
            |          |\          *
            |         *|:\       *
            |        *:|::\    *
            |      *:::|:::*
            |   *:::::*|   |\
          --*----------+---+-\-------
            |             1  \  y = -2x+6
    We need two integrals . . .

    The line intersects the parabolas at: $\displaystyle x = \tfrac{3}{4}\,\text{ and }\,x = 1$


    The first area is between the two parabolas:

    . . $\displaystyle A_1 \;=\;\int^{\frac{3}{4}}_0\bigg[8x^2-4x^2\bigg]\,dx$


    The second is between the line and the lower parabola:

    . . $\displaystyle A_2 \;=\;\int^1_{\frac{3}{4}}\bigg[(-2x + 6) - 4x^2\bigg]\,dx$


    Got it?

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  3. #3
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    Quote Originally Posted by silencecloak View Post
    $\displaystyle y = 4x^2$
    $\displaystyle y = 8x^2$
    $\displaystyle 2x+y=6$

    x > or equal to 0

    I have found the intersection points of 4x^2 and 2x+y=6 to be at (-1.5,9) and (1,4) and the intersection points of 8x^2 and 2x+y=6 to be at (-1,4) and (.75,4.5)

    I set up my integral to be from -1.5 to 1 $\displaystyle 8x^2-4x^2$ = 35/6

    I also tried setting up my integral from 0 to 1 $\displaystyle 8x^2-4x^2$ = 4/3

    But...this is wrong...can anyone help plz? thx
    did you note that the problem states that $\displaystyle x \geq 0$ ?

    $\displaystyle
    A = \int_0^{\frac{3}{4}} 8x^2 - 4x^2 \, dx + \int_{\frac{3}{4}}^1 (6-2x) - 4x^2 \, dx
    $
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