$\int_a^b x^\alpha dx=\frac{{b^{\alpha+1}}-{a^{\alpha+1}}}{\alpha+1}$

Question

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?

Answer 1

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}) $.