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Math Help - Midpoint Rule

  1. #1
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    Exclamation Midpoint Rule

    Use the Midpoint Rule to approximate the integral
    S(with a= -8 and b= -2)(8x–3x^2)dx
    with n=3.

    This seems really easy and I've done problems like these before, however I seem to have forgotten.
    Can someone please start me out?
    Thanks, any input is appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by daneeyah View Post
    Use the Midpoint Rule to approximate the integral
    S(with a= -8 and b= -2)(8x3x^2)dx
    with n=3.

    This seems really easy and I've done problems like these before, however I seem to have forgotten.
    Can someone please start me out?
    Thanks, any input is appreciated.
    So first we find \Delta x, that is the width of each subinterval we divide the interval into

    recall that \Delta x = \frac {b - a}n

    now, choose the midpoints in each of the subintervals, and call them x_1,~x_2,~x_3, ..., ~x_n

    the midpoint rule says: \int_a^b f(x)~dx \approx \Delta x [f(x_1) + f(x_2) + \cdots + f(x_n)]

    can you continue?
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  3. #3
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    Thanks, that clarifies a lot.

    So would my intervals be:

    -8,-5,-2

    That doesn't seem right, but once I get my intervals then it'll be simple from there.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by daneeyah View Post
    Thanks, that clarifies a lot.

    So would my intervals be:

    -8,-5,-2

    That doesn't seem right, but once I get my intervals then it'll be simple from there.
    \Delta x = \frac {b - a}n = \frac {-2 - (-8)}3 = 2

    thus we have the partitions are a = -8, -6, -4, -2 = b

    the midpoints, therefore, are: x_1 = -7, ~x_2 = -5, ~x_3 = -3

    now continue

    (did you draw a diagram of this?)
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