We need to calculate the area of the region bounded by the curve \( y^2 = x \), the line \( x = 4 \), and the x-axis. The process is as follows: Step 1: Rewrite the curve equation.The given equation \( y^2 = x \) can be expressed as:\[y = \sqrt{x}.\]This equation represents the upper half of the parabola, as the square root function yields only non-negative values. Step 2: Formulate the integral for the area.To find the area under the curve, we integrate \( y = \sqrt{x} \) with respect to \( x \) from \( x = 0 \) to \( x = 4 \). The area between the curve and the x-axis is given by the integral:\[\text{Area} = \int_0^4 \sqrt{x} \, dx.\] Step 3: Account for symmetry.The problem requires the area of the region enclosed by \( y^2 = x \) and the x-axis. Since \( y^2 = x \) defines a parabola symmetric about the x-axis, there are two equal areas: one above the x-axis (\( y = \sqrt{x} \)) and one below (\( y = -\sqrt{x} \)). The total area is therefore twice the area calculated in Step 2.Thus, the total area is:\[\text{Total Area} = 2 \int_0^4 \sqrt{x} \, dx.\] Step 4: Evaluate the integral.The integral of \( \sqrt{x} \) is:\[\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2}.\]Evaluating this definite integral from \( 0 \) to \( 4 \):\[\int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \left( 4^{3/2} - 0^{3/2} \right).\]Given that \( 4^{3/2} = 8 \), we get:\[\int_0^4 \sqrt{x} \, dx = \frac{2}{3} \times 8 = \frac{16}{3}.\]Finally, the total area is:\[\text{Total Area} = 2 \times \frac{16}{3} = \frac{32}{3}.\]This confirms the total area is represented by \( 2 \int_0^4 \sqrt{x} \, dx \).\[\boxed{2 \int_0^4 \sqrt{x} \, dx}\]
