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

The correct answer is Option (2) → \( \displaystyle\frac{1}{3} \) Sq. units

The curves meet where both equations hold. From x = 2y, \( y=\frac{x}{2} \); substituting in \( x^2=4y \) gives the points of intersection:

    \( \displaystyle x^2=2x \)

    \( \displaystyle x(x-2)=0 \)

    \( \displaystyle x=0,\ 2 \)

For 0 < x < 2 the line lies above the parabola (at x = 1, \( \frac{x}{2}=\frac{1}{2} \) and \( \frac{x^2}{4}=\frac{1}{4} \)). So

    \( \displaystyle \text{Area}=\int_0^2\left(\frac{x}{2}-\frac{x^2}{4}\right)dx \)

    \( \displaystyle =\left[\frac{x^2}{4}-\frac{x^3}{12}\right]_0^2 \)

    \( \displaystyle =1-\frac{8}{12} \)

    \( \displaystyle =\frac{1}{3} \)