Simpson rule or trapezoidal rule the function, $$f(x) = e^x $$ at 11 equidistant points on the interval [0,1].
The question asks whether the trapezoidal rule on 10 subintervals would give a better approximation than the Simpson rule on 5 subintervals on the provided function?, and whether we can be absolutely sure of our choice?.  
I've calculated the absolute error of each, and found out that the Simpson rule provides a better approximation. However, the question has more of a theoretical sense to it, and i was wondering if an answer could be provided without any calculations.

For Simpson's rule, $S(f)$, there exists a point $\xi \in [0,1]$ such that the error
$$
E(S(f),n) = \left|\int_0^1 f(x) dx - S(f) \right| = \frac{1}{180 n^4} |f^{(4)}(\xi)| \leq \frac{e}{180 n^4}.
$$
For the trapezoidal rule, T(f), the error is
$$
E(T(f),n) = \left|\int_0^1 f(x) dx - T(f) \right| = \frac{1}{12 n^2} |f^{"}(\xi)| \geq \frac{1}{12 n^2}.
$$
We thus have
$$
E(S(e^x),5) \leq 3e-05
$$
and
$$
E(T(e^x),10) \geq 8e-04,
$$
hence we can be sure that Simpson's rule will be the better one. In fact, I got $E(S(e^x),5) = 7.3415e-06$ and $E(T(e^x),10) = 0.0012$, so the difference is somewhat larger than the theoretical bounds predict.
