$y=x^2-x^3$ 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?

I am getting $\sum_{i=1}^{n}f(c_{i})\Delta x=\frac{5-12n+7n^2}{12n^2}$ so that the limit is $\frac{7}{12}$, which is the same as $\int_{-1}^{0}(x^2-x^3)dx$.
