$\int_0^1$ $(x^3-3x^2)$ dx
The definition of the integral that must be used is
$\int_a^b$ $f(x)$ dx = $\lim_{x\to\infty} \sum_{i=1}^n f(x_i)\Delta x $
where $\Delta x = \frac{b-a}{n}$ and $x_i=a+i \Delta x $
The answer is $\frac{-3}{4}$ and I am getting $\frac{-1}{2}$ .
Hint:
$$\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4},\qquad\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}.$$
Hint:-
Consider the partition, $\mathscr{P}=\{0<\frac{1}{n}<\frac{2}{n}<...<1\}$, Find the upper sum and lower sum. Upper integral is the infimum of the upper sum. lower integral is the supremum of lower sum.
