I am reading the book A garden of integrals and I came across this exercise:
Suppose that $f(x)=x^2, 0\leq x \leq 1.$ Show that $f$ is Riemann integrable on [$0,1$]. Hint: $\sum_P(x_k^2-x_{k-1}^2)\Delta x < \sum_P2x_k\Delta x\Delta x < 2\delta$.
Could someone guide me through the solution using the specific hint?
A major theorem in the theory of Riemann integration is that a function is Riemann integrable if and only if there exists a sequence of partitions with mesh going to zero such that the sequence of upper Darboux sums and lower Darboux sums both converge to the same finite number.
Take the sequence of uniform partitions, $x_{k,n}=k/n,k=0,1,\dots,n,n=1,2,\dots$. The upper Darboux sums will be the right hand sums, $\sum_{k=1}^n f(x_{k,n})/n$. The lower Darboux sums will be the left hand sums, $\sum_{k=0}^{n-1} f(x_{k,n})/n=\sum_{k=1}^n f(x_{k-1,n})/n$. Based on the theorem, you examine the difference $\sum_{k=1}^n (f(x_{k,n})-f(x_{k-1,n}))/n$.
The goal now is to show that this difference goes to zero as $n \to \infty$. If you want to do it "by hand", do the algebra to simplify the difference and then you'll need a bound on $\sum_{k=1}^n k$ to finish the problem.