 Solution

Let $h=(b-a)/n$ and let $M_4={\max }_{x\in [a,b]}|f^{(4)}(x)|$. A composite Simpson's rule panel covers two strips, so it has width $2h$ and absolute error at mostThere are $n/2$ panels. By the triangle inequality, the absolute error of their sum is at most the sum of their absolute errors. HenceThis assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth derivative is bounded on the interval. It is the error bound for composite Simpson's rule, expressed in strip count rather than panel count.^[ref]