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    Question

    Find the area enclosed between the curves

    y=(xтИТ4)2 and y=16тИТx2over their intersection interval.
    A 64/3 Correct Answer Incorrect Answer
    B 32/3 Correct Answer Incorrect Answer
    C 128/3 Correct Answer Incorrect Answer
    D 16/3 Correct Answer Incorrect Answer

    Solution

    To find the area enclosed between the curvesy=(xтИТ4)2 and y=16тИТx2, we first need to find the points of intersection of these two curves. We set the y values equal to each other: (xтИТ4)2 = 16тИТx2 x2 +16 тАУ 8x = 16 тАУ x2 2x2 тАУ 8x = 0 2x(x-4) = 0 The solutions are x=0 and x=4. These are the limits of our integration interval. Now we need to determine which function is greater in the interval [0,4]. Let's test a point within the interval, say x=2: Fory = (xтИТ4)2, when x=2, y=(2тИТ4)2┬а= 4 Fory = 16тИТx2, when x=2, y=16тИТ(2)2 = 12 Since 12>4 at x=2, the curvey = 16тИТx2is abovey = (xтИТ4)2 in the interval [0,4]. The area enclosed between the curves is given by the integral of the difference between the upper and lower functions over the intersection interval:

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