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Math Help - Area between curves

  1. #1
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    Area between curves

    Is there something I'm doing wrong in this problem? I can't figure it out, which probably means it's blindingly obvious.

    Find the area of the region enclosed by y=\sqrt{x}, y=\frac{1}{2}x, and x=16.

    \sqrt{x}=\frac{1}{2}x at x=(0, 4)

    So \displaystyle\int^{16}_4 \frac{1}{2}x - x^{\frac{1}{2}} dx

    \left(\frac{x^2}{4} - \frac{2\sqrt{x^3}}{3}\right) \Bigg|_{4}^{16}

    \left(64 - \frac{128}{3}\right) - \left(4 - \frac{16}{3}\right)

    Which equals \frac{68}{3}
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  2. #2
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    Quote Originally Posted by Dana_Scully View Post
    Is there something I'm doing wrong in this problem? I can't figure it out, which probably means it's blindingly obvious.

    Find the area of the region enclosed by y=\sqrt{x}, y=\frac{1}{2}x, and x=16.

    \sqrt{x}=\frac{1}{2}x at x=(0, 4)

    So \displaystyle\int^{16}_4 \frac{1}{2}x - x^{\frac{1}{2}} dx

    \left(\frac{x^2}{4} - \frac{2\sqrt{x^3}}{3}\right) \Bigg|_{4}^{16}

    \left(64 - \frac{128}{3}\right) - \left(4 - \frac{16}{3}\right)

    Which equals \frac{68}{3}
    Look at the picutre.
    Attached Thumbnails Attached Thumbnails Area between curves-picture.jpg  
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  3. #3
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    Thank you! I guess I thought that it meant only the part directly bounded by x=16...I'm new to this, so thanks.
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