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Math Help - error bound for trapezoidal and simpson rule

  1. #1
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    Exclamation error bound for trapezoidal and simpson rule

    hey everyone

    The question i got was that estimate the minimum number of subintervals with an error of magnitude less than 10^-4 by the trapezoidal rule and the simpson rule for the integral( x^3 + x) from 0 to 2

    i did the trapezoidal rule

    10^-4 = 6t(2-0)^3 / (12n^2)
    f''x is greatest at t=2

    n = 282.84 which if approximated is 283
    so n = 283

    for Simpson rule i did:

    10^-4 = 0(2-0)^5 / (180n^4)
    but i get n = 0 which i think either isn't right or the simpson rule does not have an error.

    can someone tell me if i did it right and also that whenever i get f''x or f''''x, do i always have to get a t between [a,b] that gives me the greatest value for f''x or f''''x

    thanks alot
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  2. #2
    MHF Contributor
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    Re: error bound for trapezoidal and simpson rule

    You did it correctly with the Trapezoidal Rule. For Simpson's Rule, f^{(4)}(x) = 0 for all x. So, the error is zero, no matter how many subintervals you use. In other words, Simpson's Rule does not just give an approximation, it gives the actual value of the integral, so

    \begin{align*}\int_0^2(x^3+x)dx = \left.\left(\dfrac{1}{4}x^4 + \dfrac{1}{2}x^2\right) \right]_0^2 & = 6 \\ \dfrac{2-0}{6}\left[(0^3+0) + 4\left(\left(\dfrac{0+2}{2}\right)^3+\left(\dfrac{  0+2}{2}\right) \right) + (2^3+2)\right] & = \dfrac{1}{3}\left[ 4(1^3+1)+2^3+2\right] \\ & = \dfrac{18}{3} = 6\end{align*}
    Thanks from ubhutto
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