I need to find the area under: $f(x) = 64 - x^3$ in $[2,4]$.
I know the base is $\frac{2}{n}$ and the height is $64 - \left(2 + \frac{2i}{n}\right)^3$ when I distribute everything I get
$$\frac{2}{n}\cdot \left[ 64 -8 -\frac{16i}{n} -\frac{16i^2}{n^2}
-\frac{8i^3}{n^3}\right].$$
Am I having some algebra issue here?
Hint. Let $x_k=2+\frac{2i}{n}$ for $i=0,\dots,n$, then $$A:=\int_2^{4}(64-x^3)\,dx=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n} (64-x_i^3)=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^n \left(56-\frac{24i}{n}-\frac{24i^2}{n^2}-\frac{8i^3}{n^3}\right).$$ Now, by Asymptotic behaviour of sums of consecutive powers, it follows that $\sum_{i=1}^n i^k\sim \frac{n^{k+1}}{k+1}$, which implies $$A=\lim_{n\to\infty}\frac{2}{n}\cdot\left(56n-\frac{24n^2}{2n}-\frac{24n^3}{3n^2}-\frac{8n^4}{4n^3}\right)=2(56-12-8-2)=68.$$
