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

The correct answer is Option (2) → \(\displaystyle\frac{32}{3}\)

The parabola \(y^2 = x\) is symmetric about the x-axis. The line x = 4 cuts it where \(y^2 = 4\), that is at (4, 2) and (4, −2).

So the area is twice the area between \(y = \sqrt{x}\) and the x-axis from x = 0 to x = 4:

    \(\displaystyle \text{Area} = 2\int_{0}^{4} \sqrt{x}\,dx\)

    \(\displaystyle = 2\left[\frac{2}{3}x^{3/2}\right]_{0}^{4}\)

    \(\displaystyle = \frac{4}{3} \times 8 = \frac{32}{3}\) sq. units

Tip: The region lies on both sides of the x-axis. Integrating \(\sqrt{x}\) from 0 to 4 alone gives only the upper half, \(\frac{16}{3}\), which is Option (1).