The area of the region (in square units) bounded by $x=1, x=2$ and the curve $y^2 = 4x$ in the first quadrant is 

$\frac{4}{3}[2\sqrt{2}-1]$

The correct answer is Option (1) → $\frac{4}{3}[2\sqrt{2}-1]$

$y^2 = 4x \Rightarrow y = 2\sqrt{x}$ (first quadrant)

Area $= \displaystyle \int_{1}^{2} y\,dx = \int_{1}^{2} 2\sqrt{x}\,dx$

$= 2 \int_{1}^{2} x^{1/2}\,dx$

$= 2\left[\frac{2}{3}x^{3/2}\right]_{1}^{2}$

$= \frac{4}{3}\left(2^{3/2} - 1^{3/2}\right)$

$= \frac{4}{3}(2\sqrt{2} - 1)$

$\frac{4}{3}(2\sqrt{2} - 1)$