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 $$\sum_{i=1}^n c = cn $$ $$\sum_{i=1}^n f+g =\sum_{i=1}^nf + \sum_{i=1}^ng $$ $$\sum_{i=1}^n i = \frac{n(n+1)}{2} $$ $$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$$