Obtaining $\int_0^2\big(-\frac{x^2}4-5 \big)dx$ by a Riemann sum

Question

EDIT: $R_n$ is the right endpoint of the Riemann sum

Calculate $R_n$ for the function $$f(x) = -\frac{x^2}4-5$$ on interval [0,2]

What I've done so far

$ \Delta x = \frac 2n$ and $f(x_i^*) = \frac{2i}n $

Which leads to $$ \sum_{i=1}^n f(x_i^*)\ \Delta x$$

And once you put the numbers in

$$ \sum_{i=1}^n\left( -\frac{\left(\frac{2i}n\right)^2}4-5\right) \cdot \frac 2n $$

Which leads to my issue of expanding and getting an $i$ or $i^2$ alone.

Answer 1

You have $$ \sum_{i=1}^n f(x_i^*)\ \Delta x=-\frac {2}{n^3} \sum_{i=1}^n i^2- \frac {10}{n}\sum_{i=1}^n 1 $$ then use $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ which is proved here to get, as $n \to \infty$, $$ \sum_{i=1}^n f(x_i^*)\ \Delta x \to -\frac{32}3 $$