# 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.
$f(x)=x^2+x=x(x+1)$
$\int_{-2}^{0}(x^2+x)dx, \triangle x=\frac{2}{n}, x_i=(\frac{2i}{n}-2)$
$\lim_{n\to\infty}\sum_{i=1}^{n}f((\frac{2i}{n}-2))*\frac{2}{n}=\lim_{n\to\infty}\frac{2}{n}*\sum_ {i=1}^{n}f((\frac{2i}{n}-2))$
$=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\frac{ 2i}{n}-2)(\frac{2i}{n}-1)=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\fra c{4i^2}{n^2}-\frac{6i}{n}+2)$
next I split the sums and distribute
$=\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i+\frac{4}{n}]=\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i]+\lim_{n\to\infty}\frac{4}{n}$
4/n goes to zero, then I substitute in for each of the sums using formulas:
$=\lim_{n\to\infty}\frac{8}{n^3}*\frac{n(n+1)(2n+1) }{6}-\frac{12}{n^2}*\frac{n(n+1)}{2}$
$}=\lim_{n\to\infty}\frac{4}{3}*\frac{2n^2+3n+1}{n^ 2}-6*(1+\frac{1}{n})$
$=\lim_{n\to\infty}\frac{4}{3}*(2+\frac{3}{n}+\frac {1}{n^2})-6*(1+\frac{1}{n})$
$=\frac{8}{3}-6=\frac{-10}{3}$

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

Where did I go wrong?

2. ## Re: Riemann Sums not summing

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

3. ## Re: Riemann Sums not summing

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