1. The problem asks for the area of the shaded region bounded by the curves $y^2 = x$, the vertical line $x=4$, and the $x$-axis (which is $y=0$). 2. We first express the region in terms of $y$ or $x$. Since $y^2 = x$, we have $y = \sqrt{x}$ (upper half) and because the region is above the $x$-axis, $y \geq 0$. 3. The bounded region lies between $x=0$ (from the curve at origin) and $x=4$ (given vertical line). 4. To find the area, integrate the height of the region (the curve $y = \sqrt{x}$) minus the $x$-axis ($y=0$) with respect to $x$ from 0 to 4: $$\text{Area} = \int_{0}^{4} \sqrt{x}\, dx$$ 5. Calculate the integral: $$\int \sqrt{x}\, dx = \int x^{1/2} dx = \frac{2}{3}x^{3/2} + C$$ 6. Evaluate the definite integral: $$\int_0^4 \sqrt{x}\, dx = \frac{2}{3} \left(4^{3/2} - 0 \right) = \frac{2}{3} \times 8 = \frac{16}{3}$$ 7. The area is therefore $\frac{16}{3}$. 8. Among the given options, the correct integral representation for this area is option (D): $\int_0^4 \sqrt{x}\, dx$. Final answer: **(D)** $\int_0^4 \sqrt{x}\, dx$ with the area being $\frac{16}{3}$.