The area enclosed between y^2=4x , x=1 and x=4 in first quadrant is :

The correct answer is Option (1) → \( \displaystyle\frac{28}{3} \) Sq.unit

In the first quadrant the parabola is \( y=2\sqrt{x} \). The area between the curve and the x-axis from x = 1 to x = 4 is

    \( \displaystyle \text{Area}=\int_{1}^{4}2\sqrt{x}\,dx \)

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

    \( \displaystyle =\frac{4}{3}(8-1) \)

    \( \displaystyle =\frac{28}{3}\ \text{sq. units} \)