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Let $f:[a,b] to mathbb R$ with $f(x) = x^2$, show $int_a^bf(x)dx = frac13(b^3-a^3)$

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 $|sum_i=1^nf(t_i)triangle_ix - frac13(b^3-a^3)| = |sum_1^nt_i^2triangle_ix - frac13sum_1^nx_i^3-x_i-1^3| \
= |sum_i=1^n[t_i^2 - frac13(x_i^2 + x_ix_i-1 + x_i-1^2)] triangle_ix|$

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
$$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6tag1 $$
holds. That can be done in many ways, for instance by induction or by the hockey-stick identity:
$$ sum_k=1^nk^2 = sum_k=1^nleft[2binomk2+binomk1right] = 2binomn+13+binomn+12.tag2$$
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
$$ frac1Nsum_k=1^Nleft(frackNright)^2,qquad frac1Nsum_k=0^N-1left(frackNright)^2 tag3 $$
and the limit of both, as $Nto +infty$, equals $frac13$ by $(1)$. At last,
$$ int_a^bx^2,dx = int_0^bx^2,dx - int_0^ax^2,dx = (b^3-a^3)int_0^1x^2,dxtag4 $$
proves the claim.

Addendum. With a geometric flavour (i.e. integrating along sections), $int_a^bx^2,dx$, if $0<a<b$, is just $frac1pi$ 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
$$ V = fracpi h3(R^2+Rr+r^2) $$
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 $frac13hcdot A_textbase$. 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 $frac16V_textcube$ by symmetry.

$f(x) = x^2$ is continuous everywhere, and hence Riemann-integrable. QED.

Assume $0leq a<b$.(or $a<bleq 0)$ (For the case $a<0<b$, we may come up with the equation $int_[a,b] f = int_[a,0] f + int_[0,b] f $.)

Following the notation above, we may observe $x_i-1^2 < x_ix_i-1 < x_i^2$ for each $i=1,cdots,n$ so that $$x_i-1^2 < frac13(x_i-1^2+x_i-1x_i+x_i^2)< x_i^2$$

with $x_i-1^2 leq t_i^2 leq x_i^2$ obviously.( for $a<bleq0$, "$<$" would be "$>$")

Then, $$|t_i^2-frac13(x_i-1^2+x_i-1x_i+x_i^2)|<|x_i^2-x_i-1^2|=x_i^2-x_i-1^2$$($-x_i^2+x_i-1^2$ for $a<bleq0$ )

Consequently, if we take $d=d([a,b])$ as $maxDelta_1 x,cdots,Delta_n x$,

the last term you wrote can be bounded as follows.

$$left|sum_i=1^n left(t_i^2-frac13(x_i^2+x_i-1x_i+x_i-1^2)right)Delta_i xright| leq sum_i=1^n left|t_i^2-frac13(x_i^2+x_i-1x_i+x_i-1^2)right|Delta_i x $$
$$leq d sum_i=1^n left|t_i^2-frac13(x_i^2+x_i-1x_i+x_i-1^2)right| leq d sum_i=1^n left| x_i^2 -x_i-1^2right| = d|b^2-a^2|$$

As a result, $sum_i=1^n f(t_i)Delta_i x rightarrow frac13(b^3-a^3)$ as
$drightarrow0$, regardless of the choice of $t_i$. (Rimann integrability ?)