Question

The area enclosed between y² = x and y = x is

• (A) 2/3 sq. units
• (B) 1/2 units
• (C) 1/3 units
• (D) 1/6 units

Answer 1

The Correct Option is D

Explanation:
$$A={\int }_0^1(\sqrt{x}-x)dx$$

$$\begin{array}{l}=\frac{2}{3}{\left(x^{3/2}\right)}_0^1-\frac{1}{2}{\left(x^2\right)}_0^1=\frac{2}{3}[1-0]-\frac{1}{2}[1-0]\\ =\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}\end{array}$$