Take $$x_k$$ as the left endpoint of each subinterval to find the area under the curve y = f(x) above the specified interval. f(x) = 9 − x^2; [0, 3]
What I've done so far is to consider $$\Delta x = \frac{3}{n}$$ and $$x_k=0+(k-1) \cdot \frac{3}{n}$$ My question arises when I'm going to develop the sum $$\sum_{k=1}^n (9 − ((k-1) \cdot \frac{3}{n})^2) \frac{3}{ n}$$
I know what to do to complete the problem. Could someone help me?