Answer
$\dfrac{5}{4}$
Work Step by Step
Apply formula:$\int_{p}^{q} f(x)dx=\lim\limits_{n \to \infty} \Sigma_{k=1}^nf(a+k \triangle x)$
$\int_{0}^{1} 3x-x^3 dx=\lim\limits_{n \to \infty}(\dfrac{1}{n}) \Sigma_{k=1}^n (3)(\dfrac{k}{n})-(\dfrac{k}{n})^3$
Thus, we have
$\lim\limits_{n \to \infty}[(\dfrac{3}{n^2})\dfrac{n(n+1)}{2}-(\dfrac{1}{n^4})(\dfrac{n^2(n+1)^2}{4})]=(\dfrac{3}{2}) \lim\limits_{n \to \infty}(1+\dfrac{1}{n})-\dfrac{1}{4}(1+\dfrac{1}{n})^2$
Hence, $\int_{0}^{1} 3x-x^3 dx=\dfrac{6-1}{4}=\dfrac{5}{4}$
