Question bank/JEE Main/Mathematics

Area bounded by curves y^2 = x and y = | x | is given by

A$\frac{1}{6}$B$\frac{2}{3}$C$- \frac{1}{6}$D$\frac{1}{3}$

Show answer and solution▾

[IMAGE] Point of intersection $\sqrt{x} = x$
x = 0, x = 1
$A = \int_{0}^{1}\mspace{2mu}\left( \sqrt{x} - x \right)$ dx
$= \left( \frac{x^{3/2}}{\frac{3}{2}} - \frac{x^{2}}{2} \right)_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx