We have, $y=x^{3}$ and $A(1,1)$ $\therefore \frac{d y}{d x}=3 x^{2}$ ...(i) On putting $x=1$ in E (i), we get $\frac{d y}{d x}=3(1)^{2}=3$ $\therefore$ Equation of tangent at $A(1,1)$ is $y-1=3(x-1) \Rightarrow y=3 x-2$ $\therefore$ Required area $=\int\limits_{0}^{1} x^{3} d x-\int\limits_{2 / 3}^{1}(3 x-2) d x$ $=\left[\frac{x^{4}}{4}\right]_{0}^{1}-\left[\frac{3 x^{2}}{2}-2 x\right]_{2 / 3}^{1}$ $=\frac{1}{4}-\left[\left(\frac{3}{2}-2\right)-\left(\frac{2}{3}-\frac{4}{3}\right)\right]$ $=\frac{1}{12}$ sq unit