Use the limit of a Riemann sum to evaluate integral (2x-x^2)dx? Help please?

Question

anonymous · Answer

Here is the equation:

$$\int\limits \left(\begin{matrix}3 \ 1\end{matrix}\right) (2x-x^2)dx$$

anonymous · Answer

What I have so far:

anonymous · Answer

$$\lim_{n \rightarrow \infty} \epsilon  f(2i/n + 1) \Delta x$$

anonymous · Answer

$$[2(2i/n + 1) - (2i/n + 1)^2] 2/n$$

anonymous · Answer

=$$[2+ 4i/n - (4i^2/n^2 + 4i/n + 1)] 2/n$$
=$$(4/n + 8i/n^2 - 8i^2/n^3 -8i/n - 2/n)$$

anonymous · Answer

Now I'm stuck...please help! :(

anonymous · Answer

Is this the correct next step:

$$=(\frac{ 4 }{ n } \sum_{i=n}^{n}i + \frac{ 8 }{ n^2 } \sum_{i=1}^{n}i - \frac{ 8 }{n^3 }\sum_{i=1}^{n}i^2-\frac{ 8 }{ n }\sum_{i=1}^{n}i - \frac{ 2 }{ n }\sum_{i=1}^{n}i$$

anonymous · Answer

Any help at all is appreciated!! >.<