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Math Help - Midpoint rule

  1. #1
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    Midpoint rule

    Use the Midpoint Rule to approximate the integral
    with n=3

    ***

    The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Archduke01 View Post
    Use the Midpoint Rule to approximate the integral
    with n=3

    ***

    The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?
    Midpoint rule, n should be even you can't apply it with n odd
    you can use Trapezoidal rule for odd n
    Simpson's rule same as Midpoint n even
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  3. #3
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    First, is the coefficient of x^2 supposed to be 0?

    Second, is n just the number of subintervals you're supposed to use? What's all this about the parity? Just break the integral into three parts and use the approximation \int_a^b f \,\mathrm{d}x\approx (b-a)f(\frac{a+b}2).
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  4. #4
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    Quote Originally Posted by Archduke01 View Post
    Use the Midpoint Rule to approximate the integral
    with n=3
    ***The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?
    First I am quite sure there is a typo in this post 0x^2???
    Now, with the midpoint rule n does not have to be even.
    The division points are x_n  =  - 8 + \frac{{8n}}{3};\;n = 0,1,2,3.
    The midpoints are m_k  = \frac{{x_{k - 1}  + x_k }}{2};\;k = 1,2,3.
    The midpoint sum is \sum\limits_{k = 1}^3 {\frac{8}{3}f(m_k )} .
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  5. #5
    MHF Contributor Amer's Avatar
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    My book said n even

    Theorem 4.6: let f\in C^2 [a,b] , n be even, h=\frac{(b-a)}{n+2}, and x_j = a+(j+1)h for each j=-1,0,...,n+1 . there exist a \mu \in (a,b) for which composite Midpoint rule for n+2 subintervals can be written with error term as

    \int_{a}^{b} f(x) dx = 2h \sum_{j=0}^{n/2} f(x_{2j}) + \frac{b-a}{6} h^2 f''(\mu)

    Eighth Edition
    Richard L.Burden
    J.Douglas Faires
    Youngstown State University
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  6. #6
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    That is an absolutely atrocious formula.
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