# Calculate $\int_{0}^{1} x^2 dx$ using the definition of the integral using Riemann Sums

Okay, so my Real Analysis textbook defines a definite integral as follows:

Let $$[a,b]$$ be an interval and $$f$$ a function with domain $$[a,b]$$. We say that the Riemann sums of $$f$$ tend to a limit $$l$$ as $$\text{m}(P)$$ (the mesh of $$P$$) tends to $$0$$ if, for any $$\epsilon > 0$$, there is a $$\delta > 0$$ such that if $$P$$ is any partition of $$[a,b]$$ with mesh less than $$\delta$$, then $$|R(f, P) - l| < \epsilon$$ for every choice of $$s_j \in I_j$$. The value of $$l$$ is called the Riemann integral of $$f$$ over $$[a,b]$$, denoted $$\int_{a}^{b} f(x) dx$$.

We note in the above definition that if $$P = \{x_0, x_1, \ldots, x_k\}$$ is a partition of $$[a,b]$$, $$I_j = [x_{j-1}, x_j]$$. We will let $$\Delta_j = \text{length of } I_j$$. The book emphasizes that the points of the partitions of $$[a,b]$$ need not be equally spaced.

So, given the above definitions, I'm giving the task of showing that $$\int_{0}^{1} x^2 dx = \frac{1}{3}$$

So, I basically have to show that $$|R(f, P) - \frac{1}{3}| < \epsilon$$ for every $$\epsilon > 0$$.

After giving this some thought without much success, I took a peak at the solution manual provided with my textbook, and I was surprised to discover that in the solution, the author considered an equally-spaced partition $$P$$ such that

$$R(f,P) = \sum_{j=1}^{k} (\frac{j}{k})^2\Delta_j$$ where $$\Delta_j = \frac{1-0}{k}$$

First, why can we do this? I know in most introductory calculus courses it's natural to let the Partitions of a given interval have points that are equally spaced, but the definition given in my Real Analysis book emphasized many many times that this doesn't need to be the case. In fact, I feel that by choosing a partition whose points are equally spaced, we have lost some level of generality because our definition of a definite integral requires us to consider any arbitrary partition of $$[0,1]$$, not just the partitions that make things easier for us to work with.

Second, if we are considering any arbitrary partition of $$[0,1]$$, how can we show that $$|R(f, P) - \frac{1}{3}| < \epsilon$$?

What I have so far is that

$$R(f, P) = \sum_{j=1}^{k} s_j^2\Delta_j \leq \sum_{j=1}^{k}\Delta_j \leq \sum_{j=1}^{k} \text{m}(P) = \text{m}(P)k < k\delta$$ since $$0 \leq s_j \leq 1$$ and $$\Delta_j = \text{length of } I_j \leq \text{m}(P) < \delta$$

I then need to choose a clever $$\delta$$ such that $$|R(f, P) - \frac{1}{3}| < \epsilon$$ but I don't know what I should choose. Any help on this matter would be greatly appreciated! I don't even know if I'm going in the right direction.

• If the function is continuous, the upper sum and the lower sum converge as the partition becomes sufficiently fine. For every sufficiently fine partition, the Riemann integral will converge. Nov 27 '18 at 8:46

If $$f \colon [a,b] \rightarrow \mathbb{R}$$ is continuous, then $$f$$ is Riemann-integrable over $$[a,b]$$ and then it is enough to take one shrinking partition, i.e. $$(P_n)_{n \in \mathbb{N}}$$ with $$m(P_n) \rightarrow 0$$ to evaluate the limes $$\lim_{n \rightarrow \infty} R(f,P_n)$$.
Proof: $$f$$ is uniformly contiuous on $$[a,b]$$, i.e. for any $$\varepsilon>0$$ there exists $$\delta >0$$ with $$|x-y|<\delta$$, $$x,y \in [a,b]$$ implies $$|f(x)-f(y)| < \delta$$. Now let $$P =(t_0,\ldots,t_m)$$ be any Partition with $$\mu(P)<\delta$$ and take $$n \in \mathbb{N}$$ with $$m(P_n)< \delta$$, write $$P_n=(s_0,\ldots,s_l)$$, and $$|R(f,P_n) - I| < \varepsilon$$. We have $$R(f,P_n) = \sum_{k=0}^{l+1} f(\widetilde{s}_k) (s_{k+1}-s_{k})$$ with $$\widetilde{s}_k \in [s_k,s_{k+1}]$$. We can refine the sum, such that $$m=l$$ and $$s_i =t_i$$, and similiar for $$R(f,P)$$. We may loose the property that $$\widetilde{s}_k \in [s_k,s_{k+1}]$$. However, if $$s_i \le t_k < s_{i+1}$$, then $$|\widetilde{s_i}-\widetilde{t}_k| < \delta$$. Thus the new refinement has the property that the chosen intermediate points for both partitions have distance less then $$\delta$$. Then we have \begin{align} \tag{1}|R(f,P_n) - R(f,P)| \le \sum_{k=0}^{k-1} |f(\widetilde{s_k}) - f(\widetilde{t_k})| (s_{k+1}-s_k) &< \varepsilon \sum_{k=0}^{k-1} (s_{k+1}-s_k) \\ & \le \varepsilon (b-a). \end{align} Thus, we find $$|R(f,P) -I| < \varepsilon(1+b-a)$$. The line (1) shows also (by modifing the notation), that $$(R(f,P))_{P \text{ Partition}}$$ is a Cauchy-net, i.e. convergent, because $$\mathbb{R}$$ is complete.