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Math Help - Area under curves again...

  1. #1
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    Area under curves again...

    the instruction says sketch the region enclosed by the given curves. decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. find the area of that region. I'm assuming when it says x=9, area up to x = 9
    y=\sqrt{x}
    y=1/2(x)
    x=9
    points of intersection, x = 0, x =4
    y=\sqrt{x} is on top from x = 0 to 4
    and on bottom from x = 4 to x = 9
    find area using integrals from x = 0 to x = 4 then from x = 4 to x = 9
    so area =\int_{0}^4 \sqrt{x}-\frac{1}{2}(x)dx + \int_{9}^4 \frac{1}{2}(x)-\sqrt{x}dx
    anti derivative of first integral = \frac{2}{3}x^\frac{3}{2}-\frac{1}{4}x^2
    antiderivative of 2nd integral = \frac{1}{4}x^2-\frac{2}{3}x^\frac{3}{2}
    then
    area = [F(4)-F(0)]+[F(9)-F(4)]
    When I finish all the arithmetic, i get a bunch of square roots... and the answer is 59/12
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  2. #2
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    Your first integral is correct.

    The second should be \displaystyle \int_4^9{\frac{1}{2}x - \sqrt{x}\,dx}.
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