Good afternoon!
I'm trying to solve the following problem:
"Show that:
a) given
![$f:[a,b] \rightarrow \mathbb R$ $f:[a,b] \rightarrow \mathbb R$](https://dxdy-02.korotkov.co.uk/f/d/f/5/df5d961c0d07e2fa356710fb0085370282.png)
, if

and

, then

;
b) When

, then

;
c) Use (a) and (b) to prove that, given
![$f:[a,b] \rightarrow \mathbb R$ $f:[a,b] \rightarrow \mathbb R$](https://dxdy-02.korotkov.co.uk/f/d/f/5/df5d961c0d07e2fa356710fb0085370282.png)
, whose derivative is Riemann integrable, and

, then
![$[math]f(a)+f(b)=[\frac{2}{(b-a)}] \int_a^b [f(x)+(x-m)f'(x)]dx[math]$ $[math]f(a)+f(b)=[\frac{2}{(b-a)}] \int_a^b [f(x)+(x-m)f'(x)]dx[math]$](https://dxdy-03.korotkov.co.uk/f/a/c/5/ac5e4650e9020210c832fdd7051e92fc82.png)
.
My solution:
a) Consider a partition

of [a,b], such that

, with

. The Riemann sum

of the function

over the partition

may be written as

. Then,

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) $\frac{1}{n}\sum_{i=1}^n (a+ih)=\frac{a+h+a+2h+...+a+nh}{n}=\frac{na+h(1+2+...+n)}{n}=\frac{na+nh+[h(1+2+...+(n-1)]}{n}=a+\frac{b-a}{n}+\frac{b-a}{2}\frac{n-1}{n}=a+\frac{b-a}{n}+\frac{(b-a)(n-1)}{2n}+\frac{2a(n-1)}{2n}-\frac{2a(n-1)}{2n}$=$a+\frac{b-a}{n}-\frac{a(n-1)}{n}+\frac{(b+a)(n-1)}{2n}=\frac{b}{n}+\frac{(b+a)}{2}\frac{(n-1)}{n}$ $\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^nf(a+ih)=\frac{1}{b-a}\int_a^b f(x)dx
b) $\frac{1}{n}\sum_{i=1}^n (a+ih)=\frac{a+h+a+2h+...+a+nh}{n}=\frac{na+h(1+2+...+n)}{n}=\frac{na+nh+[h(1+2+...+(n-1)]}{n}=a+\frac{b-a}{n}+\frac{b-a}{2}\frac{n-1}{n}=a+\frac{b-a}{n}+\frac{(b-a)(n-1)}{2n}+\frac{2a(n-1)}{2n}-\frac{2a(n-1)}{2n}$=$a+\frac{b-a}{n}-\frac{a(n-1)}{n}+\frac{(b+a)(n-1)}{2n}=\frac{b}{n}+\frac{(b+a)}{2}\frac{(n-1)}{n}$](https://dxdy-03.korotkov.co.uk/f/a/e/4/ae42076aacd90d4a2ad7e8dbab3be9a982.png)
.
Evidently,

when

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!