We are tasked with finding the area of the region enclosed by the curve \( y^2 = x \), the line \( x = 4 \), and the x-axis. Let's go step by step to solve the problem.
Step 1: Express the curve equation in a more useful form
The given equation is \( y^2 = x \), which can be rewritten as:
\[
y = \sqrt{x}.
\]
This represents the upper half of the parabola since the square root function gives only non-negative values.
Step 2: Set up the integral for the area
To find the area under the curve, we need to integrate \( y = \sqrt{x} \) with respect to \( x \) from \( x = 0 \) to \( x = 4 \). The area of the region between the curve and the x-axis is the integral of the function:
\[
\text{Area} = \int_0^4 \sqrt{x} \, dx.
\]
Step 3: Double the integral
However, the question asks for the area of the shaded region, which is bounded by both the curve \( y^2 = x \) and the x-axis. Since the curve \( y^2 = x \) corresponds to two symmetrical areas (one above the x-axis and one below), the total area under the curve from \( y = -\sqrt{x} \) to \( y = \sqrt{x} \) is twice the area under the curve \( y = \sqrt{x} \).
Therefore, the total area is:
\[
\text{Total Area} = 2 \int_0^4 \sqrt{x} \, dx.
\]
Step 4: Solve the integral
The integral of \( \sqrt{x} \) can be computed as:
\[
\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2}.
\]
Now, evaluate this integral from \( 0 \) to \( 4 \):
\[
\int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \left( 4^{3/2} - 0^{3/2} \right).
\]
Since \( 4^{3/2} = \), we have:
\[
\int_0^4 \sqrt{x} \, dx = \frac{2}{3} \times = \frac{16}{3}.
\]
Thus, the total area is:
\[
\text{Total Area} = 2 \times \frac{16}{3} = \frac{32}{3}.
\]
This confirms that the area is \( 2 \int_0^4 \sqrt{x} \, dx \).
\[
\boxed{2 \int_0^4 \sqrt{x} \, dx}
\]
