Let: $$ I=\int_0^3 e^{-\sin x}dx $$ $I$ has been approximated using rectangle midpoint formula $S$ (so $S(f)=H\sum_{k=1}^n\frac{x_k+x_{k-1}}{2}$), dividing interval $<0;3>$ to $150$ equal subintervals.
Prove that:
a) $|I-S|<10^{-4}$
b) $I>S$
so for part a), I used a formula that calculates rectangle midpoint formula error: $$ |I-S|=E(f)=\frac{1}{24}H^2(b-a)f''(\mu) $$ Where $H$ is the length of subintervals and $\mu\in(a,b)$
From there I got that $$ |I-S|<10^{-4}\frac{1}{2}f''(\mu) $$ so I just needed to prove that $$ f''(\mu)=e^{-\sin \mu}(-\sin^2 \mu+\sin \mu +1)<2 $$ here I subsidized $t=\sin \mu$, where $t\in <0;1>$
I got that $g(0),g(1)<2$ (where $g$ is the subsidized function $f''$)
Since $g'(t)=e^{-t}t(t-3)$, function $g$ has extremal values only is $0$ and $3$, which I already checked, as borders of the interval.
For the part b), unfortunately I have no ideas (except maybe calculating the integral analitically, and then writing a program that would calculate it's approximation, but I don't think that's the best solution)
Please verify my method for part a), and give me some tips on solving part b)
The error formula holds without the absolute value... You have that $$ I-S = \frac{1}{24}H^2 (b-a) f''(\mu) $$
If you show that $f''(\mu)>0$, you have your proof.