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)

$\frac{5}{12}$

Required area ${\int }_0^1(\sqrt{x}-x^3)dx={(\frac{2x^{3/2}}{3}-\frac{x^4}{4})}_0^1=\frac{2}{3}-\frac{1}{4}-0=\frac{5}{12}$