How can I prove this integral? I have to use the identity $b^4-a^4=(b-a)(b^3+b^2a+ba^2+a^3)$ to prove that:
$\int_b^ax^3dx=\frac{b^4-a^4}{4}$.
I know that you can just do $F(b)-F(a)$ and since the integral of $x^3$ is $\frac{x^4}{4}$ so you'd get $\frac{b^4}{4}-\frac{a^4}{4}$, but I'm not sure what method you would use to be able to use that identity.
 A: Let $f(x) = x^3$. Given a partition $P : a = x_0 < x_1 < \cdots < x_n = b$ of $[a,b]$,
$$x_{i - 1}^3 \le \frac{x_i^3 + x_i^2 x_{i-1} + x_ix_{i-1}^2 + x_{i-1}^3}{4} \le x_i^3 \quad (i = 1, 2, \ldots, n).$$ 
Thus 
$$x_{i-1}^3(x_i - x_{i-1}) \le \frac{x_i^4 - x_{i-1}^4}{4} \le x_i^3(x_i - x_{i-1}) \quad (i = 1, 2, \ldots n).$$
Taking the sum as $i$ ranges from $1$ to $n$, we obtain
$$\sum_{i = 1}^n m_i \Delta x_i \le \frac{b^4 - a^4}{4} \le \sum_{i = 1}^n M_i \Delta x_i$$
where $m_i = \min\{f(x) : x\in [x_{i-1}, x_i]\}$ and $M_i = \max\{f(x) : x\in [x_{i-1}, x_i]\}$. Since $P$ was an arbitrary partition of $[a,b]$ we conclude that $$\int_a^b x^3\, dx = \frac{b^4 - a^4}{4}.$$
A: Here it is an approach using partitions. We have:
$$\int_{a}^{b}x^3\,dx = \int_{0}^{b}x^3\,dx -\int_{0}^{a}x^3\,dx,$$
hence it is sufficient to prove that $\int_{0}^{c}x^3\,dx = \frac{c^4}{4}$ or, by setting $x=ct$,
$$\int_{0}^{1} t^3\,dt = \frac{1}{4}\tag{1}.$$
Using Riemann sums over a uniform partition we have:
$$\int_{0}^{1} t^3\,dt = \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^3 = \lim_{n\to +\infty}\frac{1}{n^4}\sum_{k=1}^{n} k^3,\tag{2}$$
but it is not difficult to prove through induction or other techniques that:
$$\sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2 = \left(\frac{n(n+1)}{2}\right)^2 =\frac{1}{4}n^4 + O(n^3)\tag{3}$$
and $(1)$ readily follows.
