$\int_a^b x^\alpha dx=\frac{{b^{\alpha+1}}-{a^{\alpha+1}}}{\alpha+1}$ I want to show that

Let $\alpha$ be a positive number. Then $\int_a^b x^\alpha dx=\frac{{b^{\alpha+1}}-{a^{\alpha+1}}}{\alpha+1}$ for $0\leq a<b$

I understand that it can be directly proved using fundamental theorems of calculs.
Instead, I should use the fact that
$(\alpha+1)s^{\alpha}<\frac{t^{\alpha+1}-s^{\alpha+1}}{t-s}<(\alpha+1)t^{\alpha}$ for $0\leq a<b$.
I think I can prove the below fact by using the mean value theorem, however I can't see how the two facts are related.
Any suggestions please?
 A: Think in terms of partitions. We have $x^\alpha$, which is an increasing function for $0\leq a < b$. Now for any partition P := {$x_0 = a, x_1, x_2,....., x_{n-1}, x_n = b$},  We have the upper sum for $P$,
$U(P, x^\alpha) = \sum_{i=1}^n x_{i}^\alpha(x_i - x_{i-1})$
since the maximum value of $x^\alpha$ in $[x_{i-1},x_i]$ is $x_i^\alpha$ .
And similarly the lower sum for $P$ is,
$L(P, x^\alpha) = \sum_{i=1}^n x_{i-1}^\alpha(x_i - x_{i-1})$
Now by proceeding by your hint which you can indeed prove by MVT,
$(\alpha+1)x_{i-1}^{\alpha}<\frac{x_{i}^{\alpha+1}-x_{i-1}^{\alpha+1}}{x_{i}-x_{i-1}}<(\alpha+1)x_{i}^{\alpha}$ for $0\leq x_{i-1}<x_{i}$.
We can rewrite it as
$x_{i-1}^{\alpha}(x_{i}-x_{i-1})<\frac{1}{\alpha+1}(x_{i}^{\alpha+1}-x_{i-1}^{\alpha+1})<x_{i}^{\alpha}(x_{i}-x_{i-1})$ for $0\leq x_{i-1}<x_{i}$.
Summing these up, we get $\forall$ P,
$L(P,x^\alpha) < \frac{1}{\alpha+1}(b^{\alpha+1}-a^{\alpha+1}) < U(P,x^\alpha)$
Now since $x^\alpha$ is integrable, hence by the squeeze theorem or whatever version of it you are familiar with,
We have,
$\int_a^b x^\alpha = \frac{1}{\alpha+1}(b^{\alpha+1}-a^{\alpha+1}) $.
