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Math Help - Simpson Rule

  1. #1
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    Simpson Rule

    Lets say the interval is [-4,4]
    I am confused to choose the correct formula in given situations.
    Can you help please.

     \int_{a}^{b} f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right]
    AND
    \int_a^b f(x) \, dx\approx <br />
\frac{h}{3}\bigg[f(x_0)+2\sum_{j=1}^{n/2-1}f(x_{2j})+<br />
4\sum_{j=1}^{n/2}f(x_{2j-1})+f(x_n)

    I think the aobve is used for n+1 nodes but I am not sure how to get the nodes. Is the original interval [-4,4] has zero nodes?
    Thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    Lets say the interval is [-4,4]
    I am confused to choose the correct formula in given situations.
    Can you help please.

     \int_{a}^{b} f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right]
    AND
    \int_a^b f(x) \, dx\approx <br />
\frac{h}{3}\bigg[f(x_0)+2\sum_{j=1}^{n/2-1}f(x_{2j})+<br />
4\sum_{j=1}^{n/2}f(x_{2j-1})+f(x_n)

    I think the aobve is used for n+1 nodes but I am not sure how to get the nodes. Is the original interval [-4,4] has zero nodes?
    Thanks
    The first is Simpson's rule with the interval divided into exactly two subintervals with the "break" at the midpoint: x_0= a, x_1= \frac{a+ b}/2, and x_1= b. The second is for any odd number of points (even number of subintervals). To find x_i, take h= \frac{b- a}{n-1}. Then x_i= a+ hi.
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