Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $A = \int_a^b [f(x) - g(x)] \, dx$.
• Intersection points: To find the intersection points of two curves, we set their equations equal to each other and solve for $x$. Then, we substitute the $x$ values back into either equation to find the corresponding $y$ values.
• Integration: Basic integration formulas, particularly for power functions: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Sketch the region and identify the bounding curves.

We are given the inequalities $x \ge 0$, $y \ge 0$, $y \ge x - 2$, and $y \le \sqrt{x}$. We want to find the area of the region defined by these inequalities. We have the curves $y = \sqrt{x}$ and $y = x - 2$. Since $x \ge 0$ and $y \ge 0$, we are only concerned with the first quadrant.

Step 2: Find the intersection points.

First, we find where $y = \sqrt{x}$ intersects the $x$-axis ($y=0$). This occurs at $x=0$. So, $(0,0)$ is one intersection point.

Next, we find where $y = x - 2$ intersects the $x$-axis ($y=0$). This occurs at $x=2$. So, $(2,0)$ is another intersection point.

Now, we find the intersection of $y = \sqrt{x}$ and $y = x - 2$. Setting them equal, we have $\sqrt{x} = x - 2$. Squaring both sides, we get $x = (x - 2)^2 = x^2 - 4x + 4$. Rearranging, we have $x^2 - 5x + 4 = 0$. Factoring, we have $(x - 1)(x - 4) = 0$. So, $x = 1$ or $x = 4$.

If $x = 1$, then $y = \sqrt{1} = 1$ and $y = 1 - 2 = -1$. Since we squared the equation, we need to check if these are valid solutions. Since $y = -1$ doesn't satisfy $y = \sqrt{x}$, $x=1$ is an extraneous solution.

If $x = 4$, then $y = \sqrt{4} = 2$ and $y = 4 - 2 = 2$. So, $(4, 2)$ is a valid intersection point.

Therefore, the curves $y = \sqrt{x}$ and $y = x - 2$ intersect at $(4, 2)$.

Step 3: Set up the integral(s).

From $x = 0$ to $x = 4$, we have $\sqrt{x} \ge x - 2$. However, from $x=0$ to $x=2$, $x-2$ is negative, and we want the area where $y \ge 0$. Thus, we have two regions to consider.

The first region is from $x=0$ to $x=2$, where the area is bounded by $y = \sqrt{x}$ and $y = 0$. The area is given by $\int_0^2 \sqrt{x} \, dx$. The second region is from $x=2$ to $x=4$, where the area is bounded by $y = \sqrt{x}$ and $y = x - 2$. The area is given by $\int_2^4 (\sqrt{x} - (x - 2)) \, dx$.

Therefore, the total area is $A = \int_0^2 \sqrt{x} \, dx + \int_2^4 (\sqrt{x} - (x - 2)) \, dx$

Step 4: Evaluate the integral(s).

First integral: $\int_0^2 \sqrt{x} \, dx = \int_0^2 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^2 = \frac{2}{3} (2^{3/2}) - 0 = \frac{2}{3} (2\sqrt{2}) = \frac{4\sqrt{2}}{3}$

Second integral: $\int_2^4 (\sqrt{x} - (x - 2)) \, dx = \int_2^4 (x^{1/2} - x + 2) \, dx = \left[ \frac{2}{3} x^{3/2} - \frac{1}{2} x^2 + 2x \right]_2^4$ $= \left( \frac{2}{3} (4^{3/2}) - \frac{1}{2} (4^2) + 2(4) \right) - \left( \frac{2}{3} (2^{3/2}) - \frac{1}{2} (2^2) + 2(2) \right)$ $= \left( \frac{2}{3} (8) - 8 + 8 \right) - \left( \frac{2}{3} (2\sqrt{2}) - 2 + 4 \right) = \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2 = \frac{16}{3} - \frac{4\sqrt{2}}{3} - \frac{6}{3} = \frac{10}{3} - \frac{4\sqrt{2}}{3}$

Step 5: Add the areas.

The total area is: $A = \frac{4\sqrt{2}}{3} + \frac{10}{3} - \frac{4\sqrt{2}}{3} = \frac{10}{3}$

However, this answer doesn't match the correct answer. Let's re-examine the problem and our steps. The mistake is that the first region is bounded by $y = \sqrt{x}$ and $y=x-2$ from $x=0$ to $x=4$, but $x-2 < 0$ from $x=0$ to $x=2$. The region we're interested in is bounded by $y=\sqrt{x}$, $y=x-2$, $x \ge 0$ and $y \ge 0$. So we integrate from $x=0$ to $x=4$. From $x=0$ to $x=2$, the area is between $y=\sqrt{x}$ and $y=0$. From $x=2$ to $x=4$, the area is between $y=\sqrt{x}$ and $y=x-2$. Thus,

$A = \int_0^4 \sqrt{x} dx - \int_2^4 (x-2) dx$ $= \left[\frac{2}{3}x^{3/2}\right]_0^4 - \left[\frac{x^2}{2} - 2x\right]_2^4 = \frac{2}{3} (8) - \left[(8-8) - (2-4)\right] = \frac{16}{3} - (0 - (-2)) = \frac{16}{3} - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3}$

Still incorrect. The correct approach is to calculate the area between $y=\sqrt{x}$ and $y=x-2$ from the intersection points. We know the intersection point is at $x=4$. The lower bound is when $x-2=0$ or $x=2$ and $y=\sqrt{x}$ intersects $y=0$ at $x=0$. The area is then $A = \int_0^4 \sqrt{x} dx - \int_2^4 (x-2) dx = \frac{16}{3} - \left[ \frac{x^2}{2} - 2x \right]_2^4 = \frac{16}{3} - \left[ (8-8) - (2-4) \right] = \frac{16}{3} - [0 - (-2)] = \frac{16}{3} - 2 = \frac{10}{3}$

This is still not the correct answer. Let's rethink the approach. The region is bounded by $y \ge x-2$, $y \le \sqrt{x}$, $x \ge 0$, and $y \ge 0$. The intersection is at $(4,2)$. The area is $\int_0^4 \sqrt{x} dx - \int_2^4 (x-2) dx = \frac{16}{3} - \left[\frac{x^2}{2} - 2x\right]_2^4 = \frac{16}{3} - \left[ (8-8) - (2-4) \right] = \frac{16}{3} - 2 = \frac{10}{3}$

Let's try something else: $\int_0^2 \sqrt{x} dx + \int_2^4 (\sqrt{x} - (x-2)) dx = \left[\frac{2}{3} x^{3/2}\right]_0^2 + \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} + 2x \right]_2^4 = \frac{2}{3} 2\sqrt{2} + \left[ (\frac{16}{3} - 8 + 8) - (\frac{4\sqrt{2}}{3} - 2 + 4) \right] = \frac{4\sqrt{2}}{3} + \frac{16}{3} - \frac{4\sqrt{2}}{3} - 2 = \frac{10}{3}$

Still $\frac{10}{3}$. Let's try integrating with respect to y. The curves are $x = y^2$ and $x = y+2$. They intersect at $y=2$. $\int_0^2 (y+2 - y^2) dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_0^2 = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18-8}{3} = \frac{10}{3}$

This also doesn't give the correct answer.

Let's try to find the area bounded by $y = \sqrt{x}, y=x-2$ and $y=0$. The area is $\int_0^2 \sqrt{x} dx - \int_0^2 (x-2) dx = \frac{4\sqrt{2}}{3} - \left[ \frac{x^2}{2} - 2x \right]_0^2 = \frac{4\sqrt{2}}{3} - [2-4] = \frac{4\sqrt{2}}{3} + 2$

It seems the provided answer is incorrect, or I am missing something. The area is $\frac{10}{3}$.

Common Mistakes & Tips

• Extraneous solutions: Squaring equations can introduce extraneous solutions, so always check your solutions in the original equation.
• Choosing the right order of integration: Sometimes, integrating with respect to $y$ instead of $x$ can simplify the problem, or vice versa.
• Sketching the region: A clear sketch is crucial for understanding the bounds of integration and which function is on top.

Summary

We are given the region defined by $x \ge 0$, $y \ge 0$, $y \ge x - 2$, and $y \le \sqrt{x}$. We found the intersection points of the curves $y = \sqrt{x}$ and $y = x - 2$. We then set up and evaluated the integral to find the area of the region. We arrived at the answer of $\frac{10}{3}$.

Final Answer

The final answer is \boxed{\frac{10}{3}}, which corresponds to option (C). However, the given correct answer is (A). It appears there is an error in the problem or the given answer. The correct area according to my calculations is $\frac{10}{3}$.