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Math Help - Midpoint Rule Help

  1. #1
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    Exclamation Midpoint Rule Help

    Using the midpoint rule, find the area with two rectangles and 4 rectangles:
    Problem: f(x) = x^3 btwn x = 0 and x = 1
    My Process:
    i) Two rectangles:
    Δ x = (b-a)/n = (1-0)/2 = 1/2
    A 1/2 [ f (1/4) + f (3/4) ]
    A 1/2 [ (1/4)^3 + (3/4)^3 ] = . 21875
    ii) Four rectangles:
    Δ x = (b-a)/n = (1-0)/4 = 1/4
    A 1/4 [ f (1/8) + f (3/8) + f (5/8) + f (7/8) ]
    A 1/4 [ (1/8)^3 + (3/8)^3 + (5/8)^3 + (7/8)^3 ] = .2422
    Is this right because when I tried doing the same process for the next problem, we got a different answer.
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  2. #2
    Junior Member slider142's Avatar
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    That is the correct sum you should get by breaking the area into a set of rectangles with the same base, and whose height is the height of the function at the midpoint of the base.
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  3. #3
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    Quote Originally Posted by slider142 View Post
    That is the correct sum you should get by breaking the area into a set of rectangles with the same base, and whose height is the height of the function at the midpoint of the base.
    So how come on question # 7,

    which asks f(x) = 1/x between x = 1 and x = 5

    i) two rectangles:

    Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

    A approx. = 2 [f(1) + f(3)] = 2 [ 1 + 1/3] = 2.66666....

    The way the solution manual did it was...


    Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

    Delta x = 1/2 x 2 + 1/4 x 2....So confused at this step???
    = 2 [1/2 + 1/4]
    = 2 [ 3/4]
    = 3/2
    What's going on?
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  4. #4
    Junior Member slider142's Avatar
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    Quote Originally Posted by Nimmy View Post
    So how come on question # 7,

    which asks f(x) = 1/x between x = 1 and x = 5

    i) two rectangles:

    Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

    A approx. = 2 [f(1) + f(3)] = 2 [ 1 + 1/3] = 2.66666....
    This is not correct, as 1 is not the midpoint of the base of the first rectangle. The base of the first rectangle is the interval [1, 1 + Deltax] = [1, 3]. The midpoint is thus (1 + 3)/2 = 2.
    Look again at the first problem. The midpoints are correct there.
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  5. #5
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    midpoint Riemann sum, two subintervals.

    \int_1^5 \frac{1}{x} \, dx \approx \frac{5-1}{2}\left[f(2) + f(4)\right] = 2\left(\frac{1}{2} + \frac{1}{4}\right)
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