TITLE: Calculating integral using summation notation
QUESTION [5 upvotes]: I'm trying to understand $$\lim_{n\to\infty}\frac{3}{n}\sum_{k=1}^{n} \left(\left(\frac{3k}{n}\right)^2- \left(\frac{3k}{n}\right) \right).$$ I believe we can take $a=0, b=3$ and so this is equivalent to $$\int_{0}^{3} (x^2-x) \, dx.$$ Is this correct?

REPLY [4 votes]: If $f$ is Riemann integrable at $[a,b]$ then
$$\lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^nf(a+k\frac{b-a}{n})=\int_a^bf(x)dx$$
in your case
$$a=0, \; b=3,\; f(x)=x^2-x$$
and
$$\int_0^3f=\frac 92$$