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 p equals

• A. 8/3
• B. 16/3
• C. 2
• D. 4/3

Answer 1

Answer: (D)

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
$=\underset{0}{\overset{1}{\int }}\left(x^2-x^3\right)dx+\underset{1}{\overset{p}{\int }}\left(x^3-x^2\right)dx$
$\frac{1}{6}=\frac{x^3}{3}-\frac{x^4}{4}{\left|{}_0^{{}^1}+\frac{x^4}{4}-\frac{x^3}{3}\right|}_1^p$
$\Rightarrow \frac{1}{6}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{p^4}{4}-\frac{p^3}{3}-\frac{1}{4}+\frac{1}{3}\right)$
$\Rightarrow \frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=\frac{3p^4-4p^3}{12}$
$\Rightarrow \frac{p^3\left(3p-4\right)}{12}=0\Rightarrow p^3\left(3p-4\right)=0$
$\Rightarrow p=0$ or $\frac{4}{3}$
Since, it is given that p > 1
$\therefore$ p can not be zero.
Hence , $p=\frac{4}{3}$