Using Riemann integral

Question

Having trouble with this problem. Find the exact value of the definite integral of $$f(x)=2-x^3$$ on the interval [0,4] by calculating $$\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x$$ using equal subintervals

Thank you

Answer 1

The Riemann sums with a partition

$$\left\{0=x_0<x_1=\frac4n<x_2=\frac8n<\ldots<x_k=\frac{4k}n<\ldots<x_n=\frac{4n}n=4\right\}\;$$

and the given function on the right extreme points of each interval are

$$\lim_{n\to\infty}\sum_{k=1}^nf\left(\frac{4k}n\right)(x_k-x_{k-1})=\lim_{n\to\infty}\sum_{k=1}^n\left(2-\left(\frac{4k}n\right)^3\right)\frac4n=$$$${}$$

$$=\lim_{n\to\infty}\frac4n\sum_{k=1}^n\left[2-\frac{64}{n^3}k^3\right]=\lim_{n\to\infty}\frac4n\left[2n-\frac{64}{n^3}\frac{n^2(n+1)^2}4\right]=$$$${}$$

$$=\lim_{n\to\infty}4\left[2-16\frac{n^2(n+1)^2}{n^4}\right]=4(2-16)=-56$$

Answer 2

Hint. One may recall that $$ \sum_{i=0}^ni^3=\left(\sum_{i=0}^ni \right)^2=\frac{n^2(n+1)^2}4. $$