Answer
$0$
Work Step by Step
Use formula such as: $\int_{p}^{q} f(x)dx=\lim\limits_{n \to \infty} \Sigma_{k=1}^nf(a+k \triangle x)$
$\int_{-1}^{1} x^3 dx=\lim\limits_{n \to \infty}(\dfrac{2}{n}) \Sigma_{k=1}^n (-1+\dfrac{2k}{n})^3=\lim\limits_{n \to \infty}(\dfrac{2}{n})\Sigma_{k=1}^n(-1)+\dfrac{8k^3}{n^3}+\dfrac{6k}{n}-\dfrac{12k^2}{n^2}$
This implies that
$\lim\limits_{n \to \infty}(-\dfrac{2}{n})\Sigma_{k=1}^n(1)+\dfrac{16}{n^4} \Sigma_{k=1}^nk^3+\dfrac{12}{n^2}\Sigma_{k=1}^n k-\dfrac{24}{n^3}\Sigma_{k=1}^nk^2=\lim\limits_{n \to \infty}[-2+4(1+\dfrac{1}{n})^2+(6)(1+\dfrac{1}{n})-(4)(1+\dfrac{1}{n})(2+\dfrac{1}{n})]$
Thus, $\int_{-1}^{1} x^3 dx=-2+4+6-8=0$
