$y=x^2-x^3$ Area under curve I have to use the limit process to fund the area of the region between the graph of the function, and the x-axis over the given interval.
$f(x)=x^2-x^3$ in the interval of $[-1,0]$
I started the problem by solving for $\Delta x$, and got that $\Delta x=\frac{1}{n}$.  Then I got $c_i=a+i\Delta x$, and got $c_i=-1+\frac{i}{n}$
Then I plugged it into the limit process:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(f(c_i)\Delta x)$
Long Story short I got $1$ to be my answer is it correct? If not then how do I get the correct answer?

To find the area under the curve, evaluate:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(f(c_i)\Delta x)$
$\Delta x = \frac{b-a}{n} = \frac{0 - (-1)}{n} = \frac{1}{n}$
$c_i = a + i\Delta x = -1 + \frac{i}{n}$
$f(c_i) = (c_i)^2 - (c_i)^3 = \frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2$
So we have,
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n((\frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2)\Delta x)$
The $\Delta x$ is constant, so we can take it out of the summation:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(\frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2)\Delta x$
Distribute the summation.
$$\lim_\limits{n \to \infty}(\sum_\limits{i=1}^n\frac{4i^2}{n^2} - \sum_\limits{i=1}^n\frac{5i}{n} - \sum_\limits{i=1}^n\frac{i^3}{n^3} + \sum_\limits{i=1}^n2)\Delta x$$
$$\lim_\limits{n \to \infty}(\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6} - \frac{5}{n}\frac{n(n+1)}{2} - \frac{1}{n^3}\frac{n^2(n+1)^2)}{4} + 2n)(\frac{1}{n})$$
$$\lim_\limits{n \to \infty}(\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6n} - \frac{5}{n}\frac{n(n+1)}{2n} - \frac{1}{n^3}\frac{n^2(n+1)^2)}{4n} + \frac{2n}{n})$$
Evaluate the limit:
$$= \frac{8}{6} - \frac{5}{2} - \frac{1}{4} + \frac{2}{1} = \frac{7}{12}$$
I wrote alot, so I might have made a mistake, but the final answer is correct so I probabbly did not
