about series and integral - solve $\int_1^2 (x^2-x)dx$ by limit of definite integral So I have to compute $$\int_1^2 (x^2-x)dx$$ by using the Riemann sums.
I got stuck the point where I got to $$\lim_{n\to \infty} \frac{1}{n^3} \sum_{i=0}^n (i^2+in).$$
Can I split the sum? If I can then I know how to continue. so the question is:


*

*can I split the sum? why can I? because as I know for only finite sums it's ok to split, however about the infinite once I'm not sure.

*if It's not legal to split the sum, I would appreciate some hint to the correct direction of solution.


Thanks!

$$I=\int_{1}^{2}(x^2-x)\,dx = \int_{0}^{1}(z^2+z)\,dz\tag{1} $$
hence:
$$ I = \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k^2+kn}{n^2}=\lim_{n\to+\infty}\frac{1}{n^3}\sum_{k=1}^{n}\left[2\binom{k}{2}+(n+1)\binom{k}{1}\right]\tag{2} $$
and:
$$ I = \lim_{n\to +\infty}\frac{1}{n^3}\left[2\binom{n+1}{3}+(n+1)\binom{n+1}{2}\right] = \frac{1}{3}+\frac{1}{2}=\color{red}{\frac{5}{6}}.\tag{3}$$
