Area (in sq. units) of the region bounded by curves $y^2 = x$ and $x = 4$ is

The curve $y^2 = x$ is a sideways parabola opening to the right with vertex at the origin $(0,0)$.

The line $x = 4$ is a vertical line.

Since $y^2 = x$ gives $y = +\sqrt{x}$ (upper half) and $y = -\sqrt{x}$ (lower half), the parabola is symmetric about the $x$-axis.

The region runs from $x = 0$ (vertex) to $x = 4$.

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$\text{Total Area} = \int_0^4 \left[\sqrt{x} - (-\sqrt{x})\right] dx$

$= \int_0^4 2\sqrt{x} \, dx$

$= 2\int_0^4 x^{1/2} \, dx$

$= 2 \left[\dfrac{x^{3/2}}{3/2}\right]_0^4$

$= 2 \times \dfrac{2}{3}\left[x^{3/2}\right]_0^4$

$= \dfrac{4}{3}\left[x^{3/2}\right]_0^4$

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$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$

$= \dfrac{4}{3}\left[8 - 0\right]$

$= \dfrac{4}{3} \times 8$

$= \dfrac{32}{3}$ sq. units