Thread: Riemann Sums not summing

1. Riemann Sums not summing

I'm working on using Riemann sums to find definite integrals. The answers aren't coming out as what they should be. Here's what I did. Sorry for all the stuff, but I'm trying to show as many steps as possible.
$\displaystyle f(x)=x^2+x=x(x+1)$
$\displaystyle \int_{-2}^{0}(x^2+x)dx, \triangle x=\frac{2}{n}, x_i=(\frac{2i}{n}-2)$
$\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}f((\frac{2i}{n}-2))*\frac{2}{n}=\lim_{n\to\infty}\frac{2}{n}*\sum_ {i=1}^{n}f((\frac{2i}{n}-2))$
$\displaystyle =\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\frac{ 2i}{n}-2)(\frac{2i}{n}-1)=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\fra c{4i^2}{n^2}-\frac{6i}{n}+2)$
next I split the sums and distribute
$\displaystyle =\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i+\frac{4}{n}]=\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i]+\lim_{n\to\infty}\frac{4}{n}$
4/n goes to zero, then I substitute in for each of the sums using formulas:
$\displaystyle =\lim_{n\to\infty}\frac{8}{n^3}*\frac{n(n+1)(2n+1) }{6}-\frac{12}{n^2}*\frac{n(n+1)}{2}$
$\displaystyle }=\lim_{n\to\infty}\frac{4}{3}*\frac{2n^2+3n+1}{n^ 2}-6*(1+\frac{1}{n})$
$\displaystyle =\lim_{n\to\infty}\frac{4}{3}*(2+\frac{3}{n}+\frac {1}{n^2})-6*(1+\frac{1}{n})$
$\displaystyle =\frac{8}{3}-6=\frac{-10}{3}$

Integrating by using the antiderivative gives me:
$\displaystyle \int_{-2}^{0}(x^2+x)dx=(\frac{x^3}{3}+\frac{x^2}{2})]_{-2}^0=0-(\frac{-8}{3}+2)=\frac{2}{3}$

Where did I go wrong?

2. Re: Riemann Sums not summing

You have $\displaystyle \frac{4}{n}$ but it should be $\displaystyle \sum_{i=1}^n \frac{4}{n}$

3. Re: Riemann Sums not summing

Thanks, I see that I was skipping a step. I'm getting the right answer now.