Question

The area enclosed between the curves $y=x^3$ and $y=\sqrt{x}$ is (in square units)

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

Answer 1

Answer: (C)

$\frac{5}{12}$

$y=\sqrt{x}$ or $y^2=x(y\ge 0)$ and $y=x^3.$

We get points of intersection $(0,0)$ and $(1,1)$

$\therefore$ Required area $={\int }_0^1(\sqrt{x}-x^3)dx$

$={[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^4}{4}]}_0^1=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$ sq. unit.