Prove that $\int_a^b x^2 dx = \frac{b^3-a^3}{3}$ I cannot assume that the integral exists as this is part of the exercise. I'm only allowed to use the definition of the integral, which is the following:
Let $f$ be defined on $[a,b]$. The function $f$ is Riemann integrable on $[a,b]$ if there exists a number $L$ such that for all $\epsilon > 0$ there is $\delta >0$ such that $$|\sigma-L| < \epsilon$$ if $\sigma$ is any Riemann sum of $f$ over a partition $P$ of $[a,b]$ such that $||P||<\delta$. 
I'm not exactly sure how to show that the integral exists using this definition. 
 A: Let us write a Riemann sum for a partition of $[a,b]$, $a=x_0<x_1<\cdots<x_n=b$ with $x_j-x_{j-1}<\delta$. Note that we can write, using the Mean Value Theorem,
$$
x_j^3-x_{j-1}^3=3d_j^2\,(x_j-x_{j-1}).
$$
for some $d_j$ with $x_{j-1}\leq d_j\leq x_j$. 
So
$$
\frac{b^3-a^3}3=\sum_{j=1}^n\frac{x_j^3-x_{j-1}^3}3=\sum_{j=1}^nd_j^2\,(x_j-x_{j-1})
$$
 Then, for points $c_1,\ldots,c_n$ with $x_{j-1}<c_j<x_j$, consider the Riemman sum
$$
\sum_{j=1}^n c_j^2\,(x_j-x_{j-1}).
$$
Note that $c_j,d_j\in[x_{j-1},x_j]$, so $|d_j^2-c_j^2|\leq x_j^2-x_{j-1}^2$.
We have
\begin{align}
\left|\frac{b^3-a^3}3-\sum_{j=1}^n c_j^2\,(x_j-x_{j-1})\right|
&=\left|\sum_{j=1}^n (d_j^2-c_j^2)\,(x_j-x_{j-1})\right|
\leq\sum_{j=1}^n |d_j^2-c_j^2|\,(x_j-x_{j-1})\\
&\leq\,\delta\,\sum_{j=1}^n|d_j^2-c_j^2|\leq\delta\,\sum_{j=1}^n(x_j^2-x_{j-1}^2)\\
&=\delta\,(x_n^2-x_0^2)=\delta\,(b^2-a^2).
\end{align}
That is, given $\varepsilon>0$, a choice of $\delta=\varepsilon/(b^2-a^2)$ will make $(b^2-a^3)/3$ satisfy the definition. 
A: You asked for a proof using the mean-value theorem. Here is one.
Let $f(x) = x^2$, and let $F(x) = \frac{1}{3}x^3$. Let a Riemann sum $\sigma$ be given, corresponding to a partition $a = t_0 < t_1 < \dots < t_n = b$, and sample points $x_1, \dots x_n$ with $x_i \in [t_{i-1},t_i]$.
Define a function $G(x)$ on $[a,b]$ as follows. Take $p$ to be the smallest integer such that $t_p \geq x$. Then set
$$G(x) = \sum_{i = 1}^{p-1} f(x_i)(t_i - t_{i-1}) + f(x_p)(x - t_{p-1}).$$
Now $G(x)$ is continuous and has a right-hand derivative everywhere, with $G_{+}'(x)$ constant equal to $f(x_i)$ on each interval $[t_{i-1},t_i)$. 
The function $f(x)$ is uniformly continuous on $[a,b]$. Let $\delta$ be given so that $|f(x) - f(y)| \leq \epsilon/(b-a)$ whenever $|x - y| \leq \delta$. If the step of the partition is $\leq \delta$, we obtain the inequality
$$|G_{+}'(x) - f(x)| \leq \epsilon/(b-a)$$
throughout $[a,b]$. Applying the mean-value theorem (for right-hand derivatives) to the function $G(x) - F(x)$, we find that
$$\left|[G(b) - F(b)] - [G(a) - F(a)]\right| \leq \epsilon.$$
It suffices to note that $G(a) = 0$ and $G(b) = \sigma$ to conclude that
$$\left|\sigma - \frac{1}{3}(b^3 - a^3) \right| \leq \epsilon,$$
which completes the proof. 
Edit: To make this more explicit, The derivative of $f$ never exceeds $2M^2$, where $M = \max(|a|,|b|)$. So you can take $\delta = \epsilon/2(b-a)M^2$.
Also, here is the specific version of the mean value theorem that I am using. If a function $h$ on $[a,b]$ is continuous and has a right-hand derivative everywhere, with $|h_{+}'(x)| \leq K$, then $|h(b) - h(a)| \leq K(b-a)$.
A: I think it is a better manner to split the answer into two parts : Existence
of the integral, and computation. For the existence part, since $x^{2}$ is
continuous on the compact interval (a,b) so it is uniformly continuous and
it suffices to follows the general proof which is presented in many
text-book (in which the authors prove that any continuous function is
Riemann integrable on $[a,b]$.) Well, if we assume the integral exists, then
for the computation it suffices to consider a particular sequence of Riemann
sums, which necessarily converge to a limit which is the required value of
the integrable.
In a first step, assume that $a=0.$ Divide the interval $[0,b]$ into n equal
subintervals of length $\Delta _{n}=\frac{b}{n}.$ Consider the following
Riemann sum (right end-point of each subinterval)
\begin{eqnarray*}
\sum_{k=1}^{n}f(\text{right endpoint}_{k})\Delta _{n} &=&\sum_{k=1}^{n}(%
\text{right endpoint}_{k})^{2}\Delta _{n} \\
&=&\sum_{k=1}^{n}(k\Delta _{k})^{2}\Delta _{n}=\Delta
_{n}\sum_{k=1}^{n}(k\Delta _{k})^{2} \\
&=&\Delta _{n}(\Delta _{n}^{2}+2^{2}\Delta _{n}^{2}+3^{2}\Delta
_{n}^{2}+\cdots +n^{2}\Delta _{n}^{2}) \\
&=&\Delta _{n}^{3}(1^{2}+2^{2}+\cdots +n^{2}) \\
&=&\frac{b^{3}}{n^{3}}(1^{2}+2^{2}+\cdots +n^{2}) \\
&=&\frac{b^{3}}{n^{3}}(\frac{n(n+1)(2n+1)}{6}) \\
&=&\frac{b^{3}}{3}\frac{(n+1)(2n+1)}{2n^{2}}
\end{eqnarray*}
then
$$
\int_{0}^{b}x^{2}dx=\lim_{n\rightarrow \infty }\frac{b^{3}}{3}\frac{
(n+1)(2n+1)}{2n^{2}}=\frac{b^{3}}{3}.
$$
If a$\neq 0,$ it suffices to make use of the additivity property of the
integral
\begin{eqnarray*}
\int_{a}^{b}x^{2}dx &=&\int_{a}^{0}x^{2}dx+\int_{0}^{b}x^{2}dx \\
&=&-\int_{0}^{a}x^{2}dx+\int_{0}^{b}x^{2}dx \\
&=&-\frac{a^{3}}{3}+\frac{b^{3}}{3} \\
&=&\frac{b^{3}-a^{3}}{3}.
\end{eqnarray*}
