Originally Posted by

**AZach** I tried to do this to compute the left hand sum with 12 rectangles. $\displaystyle \sum_{i=-1}^{11} f(xi)dx $ where $\displaystyle f(xi) = 2(-1+\frac{1i}{2})^{3})+4 $ since $\displaystyle xi = -1 + \frac{1}{2}i $

And then I basically expanded the summation out

$\displaystyle [2(-1 - \frac{1}{2})^{3} + 4] + [2(-1 - 0)^{3} + 4] + [2(-1 + \frac{1}{2})^{3} + 4] + [2(-1 + 1)^{3} + 4] + [2(-1 + \frac{3}{2})^{3} + 4] + [2(-1 + 2)^{3} + 4] + [2(-1 + \frac{5}{2})^{3} + 4] + [2(-1 + 3)^{3} + 4] + [2(-1 + \frac{7}{2})^{3} + 4] + [2(-1 + 4)^{3} + 4] + [2(-1 + \frac{9}{2})^{3} + 4] + [2(-1 + 5)^{3} + 4] + [2(-1 + \frac{11}{2})^{3} + 4] $

I know that's a lot to look at, but I basically just plug $\displaystyle i = -1 , i = 0, i = 1 ... i = 11 $ into f(xi) ... Does this look like the proper method to find the area under the curve of the function I listed originally $\displaystyle f(x) = 2x^{3} + 4 $