The area enclosed between \(y^2=x\) and \(y=x\) is

• A

  \(\frac{2}{3}\) sq. units
• B

  \(\frac{1}{2}\) units
• C

  \(\frac{1}{3}\) units
• Correct answer
  D

  \(\frac{1}{6}\) units

Solution

\(\begin{aligned} & \text {Hints: } A=\int_0^1(\sqrt{x}-x) d x \\ & =\frac{2}{3}\left(x^{3 / 2}\right)_0^1-\frac{1}{2}\left(x^2\right)_0^1=\frac{2}{3}[1-0]-\frac{1}{2}[1-0] \\ & =\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}\end{aligned}\)

Hence, the correct answer is option: 'D'.