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