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Thread: 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 $\displaystyle \Delta x$, that is the width of each subinterval we divide the interval into

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

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

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

    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.
    $\displaystyle \Delta x = \frac {b - a}n = \frac {-2 - (-8)}3 = 2$

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

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

    now continue

    (did you draw a diagram of this?)
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