How can I demonstrate the integral of $x^n$ by definition of inf sums and sup sums? I'm undergraduate student of physics and today the teacher give us a demonstration about the definite integral of $x^2$ by definitions of sums, and I understand the most of all, but I have a doubt, how can I treat the "polynomial" $M_k \Delta x_k=(a+kh)^2)$ in due to integrate this in form of sums, I know from the  geometric intuition that:
$\sum_{k=1}^n{M_k \Delta x_k}$, and given a arbitrary partition $P={[a=x_0,x_1, ... , x_n=b}]$ and $x_0<x_1<...<x_n$ we can adjust this to $h=\frac{(b-a)}{n}$ so ... $\bar{S}(f,P)=$$\sum_{k=1}^n{(a+kh)^2 h}$, and that's my doubt how can I treat that? I'll be so grateful if someone can give me some hints of how can I do.
 A: Remember you have the binomial theorem :
$(a+kh)^n=\sum_{j=0}^n{ n\choose j} a^j (kh)^{n-j} $.
In the case  $n=2$ we have $$\begin {align} 
 \lim_{n\to \infty} \sum_{k=1}^n(a+kh)^2h=\lim_{n\to \infty}\sum_{k=1}^n(a^2+2akh+k^2h^2)h=\lim_{n\to \infty} h (\sum_{k=1}^na^2+\sum_{k=1}^n2akh+\sum_{k=1}^nk^2h^2)=\lim_{n\to \infty}h (a^2\sum_{k=1}^n1+2ah\sum_{k=1}^nk+h^2\sum_{k=1}^nk^2)=\lim_{n\to \infty}  h\cdot (a^2\cdot  n+2ah\cdot \frac {n(n+1)}2+h^2\cdot \frac{n (n+1)(2n+1)}6)=\lim_{n\to \infty}(\frac {b-a}n\cdot  a^2 n+\frac{b-a}n\cdot 2a\cdot  \frac {b-a}n\cdot \frac {n (n+1)}2+(\frac{b-a}n)^3 \cdot \frac {n (n+1) (2n+1)}6 )= \lim_{n\to \infty} ({(b-a)}\cdot a^2+a\cdot  (b-a)^2\cdot 
\frac{n+1}n+\frac {(b-a)^3 (n+1)(2n+1)}{6n^2})=(b-a)a^2+a (b-a)^2+\lim_{n\to \infty}\frac {(b-a)^3 (2n^2+3n+1)}{6n^2}=ba^2+ab^2-2a^2b+\frac{(b-a)^3}3=\frac {b^3}3-\frac {a^3}3\end {align}$$.
In the interest of intellectual honesty, I will reveal that I didn't just write this out off the top of my head. ..  I looked at the following video  first...  So the "heavy  lifting" was already done... I thought I would do it in the case where $a $ and $b$ are not given for fun, as I am learning mathjax...
