The area enclosed by the curves y=x^2, y=x^3, x=0 and x=p, where p>1, is 1 / 6. The equals

Given curves are $y=x^2$y=x2 and $y=x^3$y=x3 Also, $x=0$x=0 and $x=p, p>1$x=p,p>1 Now, intersecting point is $(1,1)$(1,1)

Required Area $$\begin{aligned} & =\int_0^1\left(x^2-x^3 d x+\int_1^p x^3\right)-x^2 d x \\ & \frac{1}{6}=\frac{x^3}{3}-\left.\frac{x^4}{4}\right|_0 ^1+\frac{x^4}{4}-\left.\frac{x^3}{3}\right|_1 ^p \\ & \Rightarrow \frac{1}{6}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{p^4}{4}-\frac{p^3}{3}-\frac{1}{4}+\frac{1}{3}\right) \\ & \Rightarrow \frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=\frac{3 p^4-4 p^3}{12} \\ & \left.\Rightarrow \frac{p^3(3 p-4}{12}\right)=0 \Rightarrow p^3(3 p-4)=0 \\ & \Rightarrow p=0 \text { or } \frac{4}{3} \\ & \text { Since, it is given that } p>1 \\ & \therefore p \text { can not be zero. } \\ & \text { Hence, } p=\frac{4}{3} \end{aligned}$$​=∫01​(x2−x3dx+∫1p​x3)−x2dx61​=3x3​−4x4​​01​+4x4​−3x3​​1p​⇒61​=(31​−41​)+(4p4​−3p3​−41​+31​)⇒61​−31​+41​+41​−31​=123p4−4p3​⇒12p3(3p−4​)=0⇒p3(3p−4)=0⇒p=0 or 34​ Since, it is given that p>1∴p can not be zero.  Hence, p=34​​