prove $f(a)+f(b)$

[Brazilian]

Good afternoon!
I'm trying to solve the following problem:
"Show that:
a) given $f:[a,b] \rightarrow \mathbb R$, if $h=\frac{(b-a)}{n}$ and $M(f;n)=\frac{1}{n}\sum_{i=1}^n f(a+ih)$, then $\lim_{n\rightarrow\infty} M(f;n)=\frac{1}{b-a}\int_a^b f(x)dx$;

b) When $n \rightarrow \infty$, then $\frac{1}{n}\sum_{i=1}^n (a+ih)= \frac{n-1}{n}\frac{a+b}{2}+\frac{b}{n} \rightarrow \frac{a+b}{2}$;


c) Use (a) and (b) to prove that, given $f:[a,b] \rightarrow \mathbb R$, whose derivative is Riemann integrable, and $m=\frac{a+b}{2}$, then $[math]f(a)+f(b)=[\frac{2}{(b-a)}] \int_a^b [f(x)+(x-m)f'(x)]dx[math]$.

My solution:

a) Consider a partition $P=\{t_1,...,t_n\}$ of [a,b], such that $t_1=a+h, t_2=a+2h,...,t_n=a+nh$, with $h=\frac{b-a}{n}$. The Riemann sum $\sum(f;P*)$ of the function $f$ over the partition $P$ may be written as $\sum(f;P*)= \sum_{i=1}^n f(a+ih)(t_i-t_{i-1})=\sum_{i=1}^n f(a+ih)h=\sum_{i=1}^n f(a+ih) \frac{b-a}{n}$. Then, $\int_a^b f(x)dx= \lim_{|P|\rightarrow 0} \sum(f;P*)= \lim_{n \rightarrow \infty} $\sum_{i=1}^n f(a+ih)\frac{b-a}{n} and $\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^nf(a+ih)=\frac{1}{b-a}\int_a^b f(x)dx b) $\frac{1}{n}\sum_{i=1}^n (a+ih)=\frac{a+h+a+2h+...+a+nh}{n}=\frac{na+h(1+2+...+n)}{n}=\frac{na+nh+[h(1+2+...+(n-1)]}{n}=a+\frac{b-a}{n}+\frac{b-a}{2}\frac{n-1}{n}=a+\frac{b-a}{n}+\frac{(b-a)(n-1)}{2n}+\frac{2a(n-1)}{2n}-\frac{2a(n-1)}{2n}$=$a+\frac{b-a}{n}-\frac{a(n-1)}{n}+\frac{(b+a)(n-1)}{2n}=\frac{b}{n}+\frac{(b+a)}{2}\frac{(n-1)}{n}$.

Evidently, $\frac{n-1}{n}\frac{a+b}{2}+\frac{b}{n} \rightarrow \frac{a+b}{2}$ when $n \rightarrow \infty$

Please, help me to find a solution for item (c).
я бразилец и изучаю русский язык, но еще плохо понимаю. То, если Вы ответите на английском, я буду благодариен!

[provincialka]

It is very strange problem. May you use something more? For example, that $((x-m)f)'=f+(x-m)f'$?

[Brazilian]

Thanks for writing. The point is that I can't see a connection between itens (a),(b) and (c).