Calculate $R_n$ for $f(x)=\frac{x^2}{4}−9$ on the interval $[0,4]$ and write your answer as a function of $n$ without any summation signs. This is the last question on my homework set, and I don't understand what's wanted. I found the correct anti-derivative and calculated the area under the curve correctly. 
At first I entered the anti-derivative, but it said $~x~$ was not defined in this context. The hint says $~x_i=\frac{4i}{n}~$ and $~Δx=\frac{4}{n}~$ . " So I tried this.
$(\frac{i^2}{n-9})(\frac{4}{n})$
Then it said that $~i~$ is not defined in this context.
I'm really confused about this question. I missed the first class and caught up by reading the book, so perhaps it was mentioned in class.
 A: Hint:
$$\sum_{i=1}^N \left(\dfrac{x_i^2}4-9\right)\Delta x=\sum_{i=1}^N\left(\dfrac14\left(\dfrac{4i}{N}\right)^2-9\right)\dfrac4N$$
$$=\dfrac{16}{N^3}\left(\sum_{i=1}^N i^2\right)-9\times4=\dfrac{16}{N^3}\dfrac{N(N+1)(2N+1)}6-36$$
A: "I need to finish the p-set in the next 100 minutes. Asking the teacher isn't an option before then. I've been able to figure out all the other questions except this one."
Then you are about to be pierced by helically threaded cone!
N.F Taussig's suggestion makes the most sense- that you are asked to find the Riemann sum dividing the interval or length 4 into n equal intervals so that each interval 4/n. Given a function f(x), we take its value at one point in each interval.  Here that is either the left endpoint, 4i/n, with i going from 0 to n-1, or the right endpoint, with i going from 1 to n.  Then the value of f at that point, times the length of the interval, f(4i/n)(4/n), is the area of a rectangle with height that value of the function and width the width of the interval.  Summing over all i approximates the "area under the curve", the integral.  It appears the function is $x^2- 9$.  If that is correct then the value at 4i/n is $16i^2/n^2- 9$, not $i^2/n- 9$.
The Riemann sum is $\sum_{i=1}^n(16i^2- 9)(4/n)= \frac{64}{n}\sum_{i=1}^n i^2- \frac{36}{n}\sum_{i=1}^n 1$.  Now it should be easy to see that the sum of "1" n times is n so that last sum is $\frac{36}{n}n= 36$.  For the first sum, you would need to know the "sum of squares" formula: $\sum_{i=1}^n i^2= \frac{n(n+1)(2n+1)}{6}$.
