I have the composite Simpson's rule as $$ \frac{h}{6}(f(a)+f(b)) + \frac{h}{3}(f(a+h)+f(a+2h)+...+f(a+(n-1)h) + \frac{2h}{3}(f(a+\frac{h}{2})+(f(a+3\frac{h}{2})+...+f(a+(2n-1)\frac{h}{2}) $$
My professor has done the proofs for the midpoint and trapezoidal rule error formula, but left it to us for Simpson.
Show that if f(t) has a continuous fourth derivative then: $$ \int_{a}^b f(x)dx$ = Q_5(f) - \frac{(b-a)^{5}}{2880n^{4}} f^{4}(\theta) $$
My proof is already over 3 pages long and I feel like there must be a simpler way. I started by using a=-1,b=1 and using $\varphi$ so $$ \int_{-1}^1 \varphi(t)dt = \frac{2}{3}(\frac{1}{2}\varphi(-1)+2\varphi(0)+\frac{1}{2}\varphi(1))-\frac{1}{90}\varphi^{4}(\theta ) $$
I start by integrating by parts 4 times. Then I compute the integral from -1 to 1 twice (for t-1,t+1) and the integral from -1 to 0 and 0 to 1. Then I add them all. and divide by 3, then use the mean value theorem. Am I on the right track? It seems so tedious, and I am not getting correct answers - but again, this could be because of the tedious nature. Thanks!