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Math Help - Setting up a Riemann sum

  1. #1
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    Setting up a Riemann sum

    Set up the Riemann sum of f(x)=x^2-2 by partitioning the interval [1,5] into equal subintervals and use the left endpoint of each subinterval for x.

    I know the riemann sum in it's general form is
    n
    Σ f(ci)delta-xi and delta-x = (b-a)/n
    i=1

    But I have trouble with figuring out the subintervals and using the left endpoint. How do I come up with the left endpoint Ci?


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  2. #2
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    The left endpoints are c_i  = a + i\Delta x\;,\;i = 0,1,2 \cdots ,n - 1.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    Set up the Riemann sum of f(x)=x^2-2 by partitioning the interval [1,5] into equal subintervals and use the left endpoint of each subinterval for x.

    I know the riemann sum in it's general form is
    n
    Σ f(ci)delta-xi and delta-x = (b-a)/n
    i=1

    But I have trouble with figuring out the subintervals and using the left endpoint. How do I come up with the left endpoint Ci?
    f(x) = x^2 - 2
    let P_n := (1=x_0, x_1 , ... , x_{n-1}, x_n =5) be a partition into n equal parts..

    we have n equal partitions so, using that \delta - x =\frac{5-1}{n}=\frac{4}{n}, that is the distance between the partitions, that is, we have
    x_1 = 1 + \frac{4}{n}
    x_2 = 1 + 2\left( {\frac{4}{n}} \right)
    x_3 = 1 + 3\left( {\frac{4}{n}} \right)
    up to
    x_{n-1} = 1 + (n-1)\left( {\frac{4}{n}} \right)
    x_n = 1 + n\left( {\frac{4}{n}} \right) = 5

    so, our c_i = x_j where j = 0, 1, ..., n-1
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