The area lying between the curves y^2=4x and y=2x is :

The correct answer is Option (3) → \(\displaystyle\frac{1}{3}\)

Points of intersection: put y = 2x in \(y^2 = 4x\):

    \(4x^2 = 4x\)
    \(4x(x - 1) = 0\)
    \(x = 0 \text{ or } x = 1\)

The curves meet at (0, 0) and (1, 2). Between them the region lies above the x-axis, where the parabola is \(y = 2\sqrt{x}\). For 0 < x < 1 the parabola is above the line (at x = 1/4: \(2\sqrt{x} = 1\) and 2x = 1/2). So

    \( \displaystyle \text{Area} = \int_{0}^{1}\left(2\sqrt{x} - 2x\right)dx \)

    \( \displaystyle = \left[\frac{4}{3}x^{3/2} - x^2\right]_{0}^{1} \)

    \( \displaystyle = \frac{4}{3} - 1 = \frac{1}{3} \)