Simpson's rule

**Makall** · April 9th 2010, 02:20 AM

The Trapezoid rule applied to $\int_0^2 \! f(x) \, dx.$ gives the value 5 and the Midpoint rule gives the value 4. What value does Simpson's rule give?

It should be 13/3 but I don't see how to get it.

**Deadstar** · April 9th 2010, 11:05 AM

Midpoint Rule => $M = (b-a)f \left (\frac{a+b}{2} \right )$

Trapezoidal Rule => $T = \frac{1}{2}(b-a)(f(a) + f(b))$

Simpsons Rule => $S = \frac{b-a}{6} \bigg{[}f(a) + 4f \left (\frac{a+b}{2} \right ) + f(b) \bigg{]} = \frac{(b-a)(f(a) + f(b))}{6} + \frac{2}{3}(b-a)f \left (\frac{a+b}{2} \right )$

$= \frac{T}{3} + \frac{2M}{3} = \frac{5}{3} + \frac{8}{3} = \frac{13}{3}$

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