I want to show that $\int_0^a x^b dx=\frac{a^{b+1}}{b+1}$. Whereby $a,b\in\mathbb{N}$. I know this isn't too hard to prove with the power rule etc however I would like to form a proof using step functions. Here is my attempt but I can't quite get there!
First defining step functions $\phi, \psi$:
$$\phi=\sum^n_{j=1}\left(\frac{a(j-1)}{n}\right)^b\chi_{(a(j-1)/n,a(j)/n)}$$
$$\psi=\sum^n_{j=1}\left(\frac{aj}{n}\right)^b\chi_{(a(j-1)/n,a(j)/n)}$$
Hence:
$$\int \phi=\sum^n_{j=1}\left(\frac{a(j-1)}{n}\right)^b\left(\frac{a(j)}{n}-\frac{(a(j-1)}{n}\right)$$
$$\int \phi=\frac{a^{b+1}}{n}\sum^n_{j=1}\left(\frac{j-1}{n}\right)^b$$
$$\int \phi=\frac{a^{b+1}}{n^{b+1}}\sum^n_{j=1}(j-1)^b=\frac{a^{b+1}}{n^{b+1}}\sum^{n-1}_{j=1}(j)^b$$
So
$$\int \psi=\frac{a^{b+1}}{n^{b+1}}\sum^n_{j=1}(j)^b$$
Hence as $n\rightarrow \infty$:
$$\int \psi-\int \phi =\frac{a^{b+1}}{n^{b+1}}\left(\sum^n_{j=1}(j)^b-\sum^{n-1}_{j=1}(j)^b\right)=\frac{a^{b+1}}{n^{b+1}}j^n\rightarrow 0$$
So I know it is Riemann integrable, however I am now struggling to get the final form I want $\frac{a^{b+1}}{b+1}$. Could anyone give me a hand finishing this off? Thanks!
Your step functions seem to have a slight error: The $a$ should be included into the power as $x = a\cdot \frac{j-1}{n}$ and not just $x=\frac{j-1}{n}$. With your choice the potential values of $x$ inside the sum of $\phi$ only runs from $0$ to $1$ and not from $0$ to $a$. Recomputing should give the desired output.