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
$ = \int\limits^{1}_{0} \left(x^{2} -x^{3}\right)dx + \int\limits^{p}_{1} \left(x^{3} -x^{2}\right) dx $
$ \frac{1}{6} = \frac{x^{3}}{3} - \frac{x^{4}}{4}\left|^{^1}_{0} + \frac{x^{4}}{4} - \frac{x^{3}}{3}\right|^{p}_{1} $
$\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} - 4 p^{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 , $p = \frac{4}{3}$