Mathematics - Area Under Curves Question with Solution | TestHub

Given curves are $y=x^2$ and $y=x^3$ Also, $x=0$ and $x=p,p>1$ Now, intersecting point is $(1,1)$
Required Area $$\begin{array}{rl}& ={\int }_0^1\left(x^2-x^{3}dx+{\int }_1^p{x}^3\right)-x^{2}dx\\ & \frac{1}{6}=\frac{x^3}{3}-{\frac{x^4}{4}|}_0^1+\frac{x^4}{4}-{\frac{x^3}{3}|}_1^p\\ & \Rightarrow \frac{1}{6}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{p^4}{4}-\frac{p^3}{3}-\frac{1}{4}+\frac{1}{3}\right)\\ & \Rightarrow \frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=\frac{3p^4-4p^3}{12}\\ & \Rightarrow \frac{p^3(3p-4}{12})=0\Rightarrow p^3(3p-4)=0\\ & \Rightarrow p=0\text{ or }\frac{4}{3}\\ & \text{ Since, it is given that }p>1\\ & \therefore p\text{ can not be zero. }\\ & \text{ Hence, }p=\frac{4}{3}\end{array}$$