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For an integral of form $\int_{-2}^2 x^2e^xdx$, calculate the Newton-Cotes quadrature and estimate the error for:

  • $n=1$ (Trapezoid rule)
  • $n=2$ (Simpson's rule)
  • $n=3$ (3/8 rule)

So I know the formulas for the errors but I don't know what I'm supposed to plug into the formulas... It's very hard to find some examples rather than raw definitions so I'm kind of stuck. Should I solve the initial integral as if it were a regular integral to move on or I'm to plug something into the formula right from the beginning?

I mean: how do I benefit from knowing that the integral is equal to $ \frac{2 (e^4-5)}{e^2}\approx13.4248$?

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First, please be more specific on a number of points: 1) are you using open or closed Newton Cotes formulae?; 2) which formulae exactly have you tried and which terms don't you understand?

Still, here is the general outline on calculating the error terms. Assume we have order $n$, and let the quadrature term be $Q_{n}(f)$. Then for the error we have

$\mathrm{err}_{n}=|\int_{a}^{b}f(x)dx-Q_{n}(f)|$

The upper bound for the error term is given in terms of

$M_{n+1}:=\mathrm{max}_{a\leq x\leq b}|f^{(n+1)}(x)|$

For example, for the Trapezoid rule we have $n=1$ and $\mathrm{err}_{1}\leq\frac{(b-a)^{3}}{12}M_{2}$.

Hope this helps.

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  • $\begingroup$ The closed ones. I've tried all four from here: en.wikipedia.org/wiki/… but I just don't even know what I should use for the variables there... $\endgroup$ – Straightfw Apr 20 '13 at 17:54
  • $\begingroup$ Also, I don't know what's the sought form of the quadrature itself - as I said, I know it's $2(e^4-5)/e^2$ but do I have to transform it to a logarithm form? In all the examples I found, the results were in a form of logarithms. Do I need such form? Why? $\endgroup$ – Straightfw Apr 20 '13 at 17:55
  • $\begingroup$ logarithms? there are no logarithms in Newton Cotes (even in Wikipedia article). What do you refer to? $\endgroup$ – ConciseAndClear Apr 20 '13 at 17:59
  • $\begingroup$ Oh, silly me. I just looked at it again and the logarithms they use are there because they're the result of the initial integral :) Sorry. However: could you please tell me what are the b and a in the formulas I've shown on Wikipedia? I believe it'd be enough to get me started. $\endgroup$ – Straightfw Apr 20 '13 at 18:03
  • $\begingroup$ Oh, they're just the bounds of the integral, I get it! Will update if I find another obstacle ;) $\endgroup$ – Straightfw Apr 20 '13 at 18:05

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