Question

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

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

Answer 1

The correct option is C

$\frac{5}{12}$

Explanation for correct option

Finding the area enclosed between the curves

Given $y=x^3$and $y=\sqrt{x}$

$$\begin{gathered} x^3=\sqrt{x} \\ \Rightarrow x^6=x^2 \\ \Rightarrow x^2(x^4-1)=0 \\ \Rightarrow x=0,1 \end{gathered}$$

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

$\begin{array}{rcl}\therefore Requiredarea& =& {\int }_0^1\left(\sqrt{x}-x^3\right)dx\\ & =& {\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{x^4}{4}\right]}_0^1\\ & =& \frac{2}{3}-\frac{1}{4}\\ & =& \frac{5}{12}\end{array}$

Hence, the correct answer is option (C).