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

$\frac{28}{3}$ sq. unit

$y^2=4 x$ so $x=1, x=4$ in 1st quadrant

so  $y =\sqrt{4 x}$

$y =2 \sqrt{x}$

so area = $\int\limits_1^4 2\sqrt{x} dx$

$=2\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_1^4$

$=2\left[\frac{2}{3} x^{\frac{3}{2}}\right]_1^4$

$=\frac{4}{3}(8-1)=\frac{28}{3}$ sq. unit