The area enclosed by the curves $y=x^2, y=x^3$, $x=0$ and $x=p$, where $p1$, is $1 / 6$. The $\mathrm{p}$ equals

Question

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

Aicharya · Accepted Answer

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{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} $

The area enclosed by the curves $y=x^2, y=x^3$, $x=0$ and $x=p$, where $p>1$, is $1 / 6$. The $\mathrm{p}$…

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