Exam

JEE Main

Year

2012

Date

12 May

Mode

Online

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2012 · 12 May · Online · Mathematics · Area Under Curves

The area enclosed by the curves y=x^2, y=x^3 , x=0 and x=p , where...

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

Step-by-step solution

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-x³ d x+\int_1^p x³\right)-x² 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 p⁴-4 p³}{12}\\ & \left.\Rightarrow \frac{p^3(3 p-4}{12}\right)=0 \Rightarrow p³(3 p-4)=0 \\ & \Rightarrow p=0 \text{\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}