✅ Manual Solution:

The curve \( y^2 = x \) is symmetric about the x-axis.
The shaded area is the area above the x-axis:

Answer:
\( \int_0^4 \sqrt{x} \, dx \)

🤖 AI Generated Solution:

Absolutely! Let's break down this problem step-by-step. **Understanding the Problem** We're asked to find the area of a specific region. This region is defined by three boundaries: 1. The curve $y^2 = x$. 2. The vertical line $x = 4$. 3. The $x$-axis (which is the line $y = 0$). **Step 1: Sketching the Region** It's crucial to visualize the region we're dealing with. * **The curve $y^2 = x$**: This equation describes a parabola that opens to the right. If we solve for $y$, we get $y = \pm \sqrt{x}$. This means the parabola is symmetric with respect to the $x$-axis. * **The line $x = 4$**: This is a vertical line passing through $x = 4$. * **The $x$-axis ($y = 0$)**: This is the horizontal axis. When you sketch these, you'll see that the region is bounded by the upper half of the parabola $y = \sqrt{x}$, the vertical line $x = 4$, and the $x$-axis. **Step 2: Determining the Limits of Integration** To find the area using integration, we need to know the range of $x$-values (or $y$-values) that define our shaded region. * The region starts at the origin (where the parabola intersects the $x$-axis), so the lower limit for $x$ is $0$. * The region ends at the vertical line $x = 4$, so the upper limit for $x$ is $4$. Thus, our integration will be with respect to $x$, from $x = 0$ to $x = 4$. **Step 3: Setting up the Integral** The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral: $$ Area = \int_{a}^{b} f(x) \, dx $$ In our case: * The function defining the upper boundary of the region is $y = \sqrt{x}$. * The limits of integration are $a = 0$ and $b = 4$. So, the integral for the area is: $$ Area = \int_{0}^{4} \sqrt{x} \, dx $$ **Step 4: Evaluating the Integral** Now, we need to solve this definite integral. * Rewrite $\sqrt{x}$ as $x^{1/2}$. $$ Area = \int_{0}^{4} x^{1/2} \, dx $$ * Use the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1}$. $$ Area = \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{0}^{4} $$ $$ Area = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} $$ $$ Area = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} $$ * Now, apply the limits of integration (upper limit minus the lower limit). $$ Area = \frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2} $$ $$ Area = \frac{2}{3} (4 \cdot \sqrt{4}) - 0 $$ $$ Area = \frac{2}{3} (4 \cdot 2) $$ $$ Area = \frac{2}{3} (8) $$ $$ Area = \frac{16}{3} $$ **Conclusion** The area of the shaded region bounded by the curve $y^2 = x$, $x = 4$, and the $x$-axis is $\frac{16}{3}$ square units. The final answer is $\boxed{\frac{16}{3}}$.