<s>[INST] How to evaluate the definite integral by the limit definition $\int_{-1}^1 x^3 dx$? Solve the definite integral by the limit definition:
$$\int_{-1}^1 x^3 dx$$
The formula:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n f(c_i)\Delta x_i$$
Get the variables:
$$\Delta x_i : \frac{b-a}{n} = \frac{1-(-1)}{n} = \frac{2}{n}$$
$$f(c_i) : a + i(\Delta x_i) = \left(-1+\frac{2i}{n}\right)$$
Now plug them into the formula I get:
$$\lim_{n\rightarrow \infty} \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3\left(\frac{2}{n}\right)$$
Take out the delta and distribute the cube:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3 = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{8i^3}{n^3}\right) $$
Expand the summation and take out constants, and properties of Simga:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[- \sum_{i=1}^n 1+ \frac{8}{n^3} \sum_{i=1}^n i^3\right] = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[-n + \frac{8}{n^3}\left( \frac{n^2(n+1)^2}{4}\right)\right]$$
Distribute the $\frac{2}{n}$ and simplify:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{16}{n^4}\left( \frac{n^2(n+1)^2}{4}\right)\right] = \lim_{n\rightarrow \infty}\left[-2 + \frac{4}{n^2}\left( \frac{n^2+2n+1}{1}\right)\right]$$
Distribute and simplify/cancel:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{4n^2}{n^2} + \frac{8n}{n^2} + \frac{4}{n^4}\right] = \lim_{n\rightarrow \infty}\left[-2 + 4 + \frac{8}{n} + \frac{4}{n^4}\right] $$
I keep getting the limit is -2 (-2+4), but the book says it's 0. Where did I go wrong?
 [/INST] When you said
"distribute the cube",
you made a mistake.
(mistake in following expansion corrected)
$(-1+2i/n)^3
=-1+3(2i/n)-3(2i/n)^2+(2i/n)^3
$.
Each term has to be considered.
Also,
note that the function is odd.
What happens if 
you take terms
symmetrically located
around the origin?
</s>