Midpoint Approximation

Derive a summation formula for midpoint approximations.

$\displaystyle \frac{(b-a)}{n}$ * $\displaystyle \sum_{i=0}^{n-1} (f(a+\frac{(b-a)}{2n}$ + $\displaystyle \frac{(b-a)*i}{n}$ ))

Plug in the function $x^{3}$ , endpoints $a$ and $b$ , and the number of rectangles $n$ .

$\displaystyle \frac{(3-1)}{50}$ * $\displaystyle \sum_{i=0}^{50-1} ((1+\frac{(3-1)}{2*50}$ + $\displaystyle \frac{(3-1)*i}{50}$ ) $^{3}$ )

Use a calculator to evaluate this sum which comes out to be $\boxed{19.9984}$