Mathematics - Area Under Curves Question with Solution | TestHub

We have, $y=x^3$ and $A(1,1)$ $\therefore$ $\frac{dy}{dx}=3x^2$ On putting $x=1$ in $\mathrm{Eq}$. $\frac{dy}{dx}=3(1{)}^2=3$ $\therefore$ Equation of tangent at $A(1,1)$ is $y-1=3(x-1)\Rightarrow y=3x-2$$\therefore$ Required area $={\int }_0^1{x}^{3}dx-{\int }_{2/3}^1(3x-2)dx$ $={\left[\frac{x^4}{4}\right]}_0^1-{\left[\frac{3x^2}{2}-2x\right]}_{2/3}^1$ $=\frac{1}{4}-\left[\left(\frac{3}{2}-2\right)-\left(\frac{2}{3}-\frac{4}{3}\right)\right]$ $=\frac{1}{12}\mathrm{sq}$ unit