Hey guys I have this problem:
Recall that ${\cal R}(f,{\cal U}_N)$ is the upper Riemann sum and ${\cal R}(f,{\cal L}_N)$ is the lower Riemann sum. Suppose $f(x)$ is continuous and increasing on $[a, b]$. Show that $${\cal R}(f,{\cal U}_N)-{\cal R}(f,{\cal L}_N)=\big(f(b)-f(a)\big)\cdot\dfrac{b-a}N.$$ Show that a similar statement is true if $f(x)$ is continuous and decreasing. Conclude that $f$ is integrable on $[a,b]$ in either case, i.e. $$\lim_{N\to\infty}\big({\cal R}(f,{\cal U}_N)-{\cal R}(f,{\cal L}_N)\big)=0.$$ For continuous function $f$ on $[a,b]$, we can break up $[a,b]$ int intervals where the function is increasing and where the function is decreasing and thus prove (using the reasoning here) that any continuous function $f$ is integrable on $[a,b]$.
I'm not 100% sure where to start.
Hints: Take $f(x)$ continuous and increasing, and consider it in the interval $[x_i, x_{i+1}]$. What is the maximum value of $f(x)$ in that interval? What is the minimum value? What is $\max(f) \cdot \Delta x_i$? What is $\min(f) \cdot \Delta x_i$? What is their difference? What are two expressions for what you get when you add all these differences over all the intervals $[x_0, x_1], \ldots, [x_{N-1}, x_N]$?