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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?

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

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

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

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

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