My Math Forum Solving Riemann Lower sum

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 May 26th, 2017, 02:53 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Solving Riemann Lower sum I'm trying to find the integral of $\displaystyle f(x)=x^{2},f:[0,1]\rightarrow \mathbb{R}$ using Riemann's criterion. I created a partition with equal intervals like this $\displaystyle P_{n}=\{0<\frac{1}{n}<\frac{2}{n}<...<\frac{n-1}{n}=1\}$ and then I started cxalculating the lower sum $\displaystyle L(f,P_{n})=f(0)\frac{1}{n}+f(\frac{1}{n})\frac{1}{ n}+...+f(\frac{n-1}{n})\frac{1}{n}$$\displaystyle =\frac{1}{n}(0+\frac{1}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}})$$\displaystyle =\frac{1^{2}+2^{2}+...+(n-1)^{2}}{n^{3}}$ I don't know how to proceed in my book it says that this is equal to: $\displaystyle \frac{(n-1)n(2n-1)}{6n^{3}}$ but I don't understand why.. Can anyone explain how we came to the conclusion that these are equal and what is the way to construct a general equation like this?
 May 26th, 2017, 04:07 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 $\displaystyle L = \left ( \sum_{j=1}^n f \left \{ 0 + \dfrac{j - 1}{n} * (1 - 0) \right \} \dfrac{1}{n} \right ) = \dfrac{1}{n} * \sum_{j=1}^n f \left ( \dfrac{j - 1}{n} \right )^2 = \dfrac{1}{n^3} * \left ( \sum_{j=1}^{n-1}j^2 \right ).$ So you got here: good work. $1^2 = 1 = \dfrac{6}{6} = \dfrac{1 * 2 * 3}{6} = \dfrac{1(1 + 1)(2 * 1 + 1)}{6}.$ $1^2 + 2^2 = 5 = \dfrac{30}{6} = \dfrac{2 * 3 * 5}{6} = \dfrac{2(2 + 1)(2 * 2 +1)}{6}.$ $1^2 + 2^2 + 3^2 = 14 = \dfrac{84}{6} = \dfrac{3 * 4 * 7}{6} = \dfrac{3(3 + 1)(2 * 3 + 1)}{6}$ In general it can be shown that $\displaystyle \sum_{j=1}^mj^2 = \dfrac{m(m + 1)(2m + 1)}{6}.$ So if m = n - 1, we get $\displaystyle L = \dfrac{1}{n^3} * \sum_{j=1}^{m=n-1}j^2 = \dfrac{1}{n^3} * \dfrac{m(m + 1)(2m - 1)}{6} = \dfrac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6n^3} \implies$ $L = \dfrac{(n - 1)n(2n - 1)}{6n^3} = \dfrac{2n^2 - 3n + 1}{6n^2} = \dfrac{1}{3} - \dfrac{1}{2n} + \dfrac{1}{6n^2}.$ Last edited by JeffM1; May 26th, 2017 at 04:18 AM.
 May 26th, 2017, 04:46 AM #3 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Ok I see what's going on. Can you please tell me how construct such equalities? For example if I want to construct an equality for $\displaystyle \sum_{j=1}^{m}\sqrt{j}$. Is there any specific way of working or do you just use your imagination and intuition and test it to see if it's correct?

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