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Math Help - Intergration Help!

  1. #1
    Member looi76's Avatar
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    Intergration Help!

    Question:
    Find the area enclosed between the curves y = x^2 + 7 and y = 2x^2 + 3.

    Attempt:

    [ y = x^2 + 7 ] \times 2
    [ y = 2x^2 + 3 ]

    2y = 2x^2 + 14
    y = 2x^2 + 3
    y = 11

    11 = 2x^2 + 3
    11 - 3 = 2x^2
    8 = 2x^2
    \frac{8}{2} = x^2
    4 = x^2
    x = \sqrt{4}
    x = 2

    y = 11 , x = 2

    y = x^2 +7

    = \frac{x^{2+1}}{2+1} +7x

    = \frac{x^3}{3} + 7x

    = (\frac{1}{3}x^3 + 7x)^2_0

    = (\frac{1}{3}\times2^3 + 7\times2) - (\frac{1}{3}\times0^3 + 7\times0)

    = 16\frac{2}{3} - 0

    = 16\frac{2}{3}

    y = 2x^2 + 3

    = \frac{2x^{2+1}}{2+1} + 3x

    = \frac{2x^3}{3} + 3x

    = (\frac{2}{3}x^3 + 3x)^2_0

    = (\frac{2}{3}\times2^3 + 3\times2) - (\frac{2}{3}\times0^3 + 3\times0)

    = 11\frac{1}{3}

    Enclosed Area = 16\frac{2}{3} - 11\frac{1}{3}
    = 5\frac{1}{3}

    Answer in the text book is 10\frac{2}{3}
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  2. #2
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    First, find the points where the two functions intersect, which is to say you have to define the "ends" of the enclosed area. Both functions are equal to y, so set them equal to one another:

    x^2 + 7 = 2x^2 + 3

    Solving that gives x= \pm 2 - remember \sqrt{4} = \pm 2 - I think this is the step you missed.

    In the interval (-2,2), x^2 + 7 has a greater value than 2x^2 + 3. In other words, x^2 + 7 will be the "top" function.

    You're going to have to integrate with respect to x as x goes from -2 to 2, as those are the "ends" of the region of intersection. To find the area enclosed between two curves, you integrate f(x) - g(x) where f(x) is the "top" function and g(x) is the "bottom" function.

    So your integral is:

    \int^2_{-2} [(x^2 + 7) - (2x^2 + 3)] dx

     =  \int_{-2}^2 (-x^2 + 4) dx

     = - \frac{1}{3}x^3 + 4x]^2_{-2}

     = (- \frac{8}{3} + 8) - ( \frac{8}{3} - 8)

     = 10 \frac{2}{3}.

    Hope this helps!
    Last edited by Mathnasium; February 29th 2008 at 12:19 AM.
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