# Basic Integration: Evaluating the Lower Sum and the Limit of the Lower Sum as $n \rightarrow \infty$

I know this is a basic integration question, but I am just beginning to learn the subject and don't fully understand it yet - which is why I don't know the answer to this question.
The question is:
For $$f(x) = x^{2}$$, divide the interval [0,2] into $$n$$ equally-wide subintervals and evaluate the lower sum and the limit of the lower sum as $$n \rightarrow \infty$$.

I know that if $$n \rightarrow \infty$$ then I am taking the limit of the Riemann Sum - or using the integral. But, how would I solve this problem step by step?

Using a Riemann Sum (With Right Endpoints):

$$\lim_{n \to \infty}(\frac{b-a}{n}\sum_{i=1}^nf(a +i(\frac{b-a}{n})))$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(i(\frac{2}{n}))^2)$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(\frac{4i^2}{n^2}))$$ Factor out the constants $$\lim_{n \to \infty}(\frac{8}{n^3}\sum_{i=1}^n(i^2))$$ Using the summation property of $$i^2$$ $$\lim_{n \to \infty}(\frac{8}{n^3}(\frac{n(n+1)(2n+1)}{6}))$$ Distribute the n $$\lim_{n \to \infty}(\frac{8}{n^3}(\frac{(2n^3+3n^2+n)}{6}))$$ $$\lim_{n \to \infty}(\frac{8(2n^3+3n^2+n)}{6n^3})$$ Because this limit tends to infinity, we look at the highest-degree n's. $$\lim_{n \to \infty}\frac{16n^3}{6n^3}$$ $$\lim_{n \to \infty}\frac{8}{3}$$ $$\frac{8}{3}$$

Summation Properties $$\sum_{i=1}^n cf = c*\sum_{i=1}^n f$$ Where c is any constant $$\sum_{i=1}^n c = cn$$ $$\sum_{i=1}^n f+g =\sum_{i=1}^nf + \sum_{i=1}^ng$$ $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$$

• What is the summation property of $i^{2}$? – Burt Jul 17 '19 at 3:29
• $\frac{n(n+1)(2n+1)}{6}$ – N. Bar Jul 17 '19 at 3:30
• OP asked for lower sum – J. W. Tanner Jul 17 '19 at 4:40
• @N.Bar - you offered to list some summation properties. If you don't mind, that would really help me out. – Burt Jul 17 '19 at 14:10

Hint:

$$\sum_{i=0}^{n-1}(i\Delta x)^2\Delta x= (\Delta x)^3\sum_{i=0}^{n-1}i^2=\left(\dfrac2n\right)^3\dfrac{(n-1)n(2n-1)}{6}$$

• Why is it $(i \Delta x)^{2}$ and not just $x^{2}$ – Burt Jul 17 '19 at 3:17
• Think of $i\Delta x$ as $x$. As $i$ ranges from $0$ to $n$, $i\Delta x$ ranges from $0$ to $2$, with $\Delta x=\dfrac 2n$ – J. W. Tanner Jul 17 '19 at 3:21
• Note: $\dfrac{(n-1)n(2n-1)}{n^3}=\dfrac{2n^3-3n^2+n}{n^3}=2-\dfrac3n+\dfrac1{n^2}$ – J. W. Tanner Jul 17 '19 at 3:26