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Thread: Can someone tell me what I did wrong in solving this problem?

  1. #1
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    Can someone tell me what I did wrong in solving this problem?

    Sketch the region enclosed by the given curves.y = 4/x, y = 16x, y =x/16, x>0
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  2. #2
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    Re: Can someone tell me what I did wrong in solving this problem?

    I assume we want to find the area of the region

    \int_0^{1/2} \left(16x-\frac{x}{16}\right) \, dx+\int_{1/2}^8 \left(\frac{4}{x}-\frac{x}{16}\right) \, dx=16 \ln  2\approx  11.0904
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  3. #3
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    Re: Can someone tell me what I did wrong in solving this problem?

    The graph of y= 4/x is a hyperbola passing through (1/2, 8), (2, 2), and (8, 1/2). y= 16x is the straight line through (0, 0) and (1/2, 8). y= x/16 is the straight line through (0, 0) and (8, 1/2). The region enclosed is the "triangle" (one curved side) with vertices (0, 0), (1/2, 8), and (8, 1/2). You should have two integrals from 0 to 1/2 and from 1/2 to 8 because, for x< 1/2, the region goes from the straight line y= x/16 up to the straight line y= 16x while for 1/2< x< 8, it goes from the straight line y= x/16 up to the hyperbola y= 4/x.
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