Question

Evaluate the integral by computing the limit of the Riemann sums for a regular partition of the interval of integration

Answer 1

\[\int\limits_{-1}^{2} (4x^{2}-3)dx \]

Answer 2

I can evaluate the limit, I just don't know how to do it with Riemanns sum

Answer 3

...

Answer 4

What do you mean left handed sum?

Answer 5

-1 0 1 2


Left handed sum

f(-1)(1)+f(0)(1)+f(1)(1)

right handed sum
f(0)(1)+f(1)(1)+f(2)(1)

Answer 6

In general, a Riemann sum looks like \[\sum_{i=0}^n f(x_i)\Delta x\]Where \(\displaystyle\Delta x=\frac{b-a}{n}\). Give me a minute to remind myself how best to evaluate these.

Answer 7

Let's take the right Riemann sum to begin with. Then we take the sum \[\sum_{i=1}^n f(x_i)\frac{b-a}{n}\]with \[x_i=-1+\frac{3i}{n}.\]So we find that \[f(x_i)=4x^2-3=4(-1+\frac{3i}{n})^2-3=\frac{36i^2}{n^2}-\frac{24i}{n}+1.\]So we really have the sum\[\frac{3}{n}\sum_{i=1}^n\left[\frac{36i^2}{n^2}-\frac{24i}{n}+1\right]\]\[=\frac{108}{n^3}\sum_{i=1}^n i^2-\frac{72}{n^2}\sum_{i=1}^n i+\frac{3}{n}\sum_{i=1}^n 1\]You can evaluate all these sums fairly easily, and then you just take the limit as \(n\to\infty\).

Answer 8

I'll be gone for a few minutes if you have questions.