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The function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Right Rectangle Approximation, using the 4 intervals between those given points.

x 4 9 11 14 15
f(x) -6 -11 -18 -21 -25

-11.545
-14
-16.273
-18.75

Please help me. I really don't know what to do.

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3 answers
  1. Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x

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  2. The x values given are the right-left edges of the rectangles. You want right-side values, so figure f(x) at

    x = 9,11,14,15

    Multiply each height by its interval's width, and the approximation is just

    5*11+2*18+2*21+1*25 = 158

    So, the average height is the area divided by the width: 158/11 = 14.36 (minus, of course) or -14

    As for y=x^3, note that the graphs intersect at x=0,1.

    So, you want 4 intervals of width 0.25, and all you do is find

    f(x) at x = 1/8, 3/8, 5/8, 7/8

    multiply each value by 0.25 and add 'em up.

    It's just a bunch of rectangles, not something scary.

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  3. You have to find the average value of the function.

    average value = [1/(b-a)] [the integral of f(x) dx]

    The integral will be solved using the right approximation.
    a and b are the domain, so a = 4 and b = 15

    average value = [1/(15 - 4)] [(-11)(5) + (-18)(2) + (-21)(3) + (-25)(1)]
    average value = 1/11 [-179]
    average value = -179/11 = -16.273

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