TITLE: Taking the integral $x^3$ using the limit definition
QUESTION [0 upvotes]: So I got as far as setting up the limit as:
$$\lim_{n\to \infty} \sum_{i = 1}^{n} \left(\frac 6n\right)\left(-3+6\frac in\right)^3$$
However, I'm ending up with $$\lim\left(\left(\frac 6n\right)(27n),n,\infty\right)$$ which is giving me $n$'s in both numerator and denominator, so as a result I am getting a non zero number. What am I doing wrong?
Please help!

REPLY [0 votes]: Actually it is the limit of the Riemann sum and one can has
$$\lim_{n\to \infty} \sum_{i = 1}^{n} \left(\frac 6n\right)\left(-3+6\frac in\right)^3=6\int_0^1(-3+6x)^3dx=0.$$
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