Question

The area enclosed between the curves $y=x^3$ and $y=\sqrt{x}$ is

• A. $\frac{5}{3}$ sq unit
• B. $\frac{5}{4}$ sq unit
• C. $\frac{5}{12}$ sq unit
• D. $\frac{12}{5}$ sq unit

Answer 1

Answer: (C)

Solving the given curves $y=\sqrt{x}$ or $y^2=x$ and $y=x^3$

we get, the points of intersection $(0,0)$ and $(1,1)$.
$\therefore$ Required area $=\underset{0}{\overset{1}{\int }}\left(\sqrt{x}-x^3\right)dx$
$={\left[\frac{x^{3/2}}{3/2}-\frac{x^4}{4}\right]}_0^1$
$=\frac{5}{12}$ sq unit