Let $f:[a,b] \to \mathbb R$ with $f(x) = x^2$, show $\int_{a}^{b}f(x)dx = \frac{1}{3}(b^{3}-a^{3})$
Here is my trial:
Let $X =(x_{0},\dots,x_{n})$ be a partition on $[a,b]$, let $t_{i} \in [x_{i-1},x_{i}], i=\{1,\dots,n\}$
Then $|\sum_{i=1}^{n}f(t_{i})\triangle_{i}x - \frac{1}{3}(b^{3}-a^{3})| = |\sum_{1}^{n}t_{i}^{2}\triangle_{i}x - \frac{1}{3}\sum_{1}^{n}x_{i}^{3}-x_{i-1}^{3}| \\
= |\sum_{i=1}^{n}[t_{i}^{2} - \frac{1}{3}(x_{i}^{2} + x_{i}x_{i-1} + x_{i-1}^{2})] \triangle_{i}x|$
But now I got stuck... can someone help me out?
I guess this answer has been given many times, but I am too lazy to find a reference, I prefer to write it again. It is useful to prove first that $$ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\tag{1} $$ holds. That can be done in many ways, for instance by induction or by the hockey-stick identity: $$ \sum_{k=1}^{n}k^2 = \sum_{k=1}^{n}\left[2\binom{k}{2}+\binom{k}{1}\right] = 2\binom{n+1}{3}+\binom{n+1}{2}.\tag{2}$$ For the function $f(x)=x^2$ over the interval $(0,1)$, the upper and lower Riemann sums for uniform partitions are given in terms of $$ \frac{1}{N}\sum_{k=1}^{N}\left(\frac{k}{N}\right)^2,\qquad \frac{1}{N}\sum_{k=0}^{N-1}\left(\frac{k}{N}\right)^2 \tag{3} $$ and the limit of both, as $N\to +\infty$, equals $\frac{1}{3}$ by $(1)$. At last, $$ \int_{a}^{b}x^2\,dx = \int_{0}^{b}x^2\,dx - \int_{0}^{a}x^2\,dx = (b^3-a^3)\int_{0}^{1}x^2\,dx\tag{4} $$ proves the claim.
Addendum. With a geometric flavour (i.e. integrating along sections), $\int_{a}^{b}x^2\,dx$, if $0<a<b$, is just $\frac{1}{\pi}$ times the volume of a frustum of a cone with height $b-a$ and radii of the bases given by $a$ and $b$. The formula $$ V = \frac{\pi h}{3}(R^2+Rr+r^2) $$ is known since the Babylonian period and can be easily proved through similarities, once we know that the volume of a cone (or pyramid) is $\frac{1}{3}h\cdot A_{\text{base}}$. Through Cavalieri's principle and the property of linear maps of proserving ratios of volumes, the whole question boils down to finding the volume of a pyramid given by the convex envelope of a face of a cube and the center of the same cube. That trivially is $\frac{1}{6}V_{\text{cube}}$ by symmetry.