Let's compute the area of the region enclosed by the given curves:

1. Equations:
   • $y = 2\sqrt{x}$
   • $y = x - 2$
   • $x + y = 2$

Step 1: Find the points of intersection

1. Intersection of $y = 2\sqrt{x}$ and $y = x - 2$: $2\sqrt{x} = x - 2$ Square both sides: $4x = (x - 2)^2$ Expand and simplify: $4x = x^2 - 4x + 4$ $x^2 - 8x + 4 = 0$ Use the quadratic formula: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{8 \pm \sqrt{64 - 16}}{2}$ $x = \frac{8 \pm \sqrt{48}}{2}$ $x = \frac{8 \pm 4\sqrt{3}}{2}$ $x = 4 \pm 2\sqrt{3}$ So the $x$-coordinates are $x_1 = 4 - 2\sqrt{3}$ and $x_2 = 4 + 2\sqrt{3}$.

   For each $x$, find $y$: $y_1 = 2\sqrt{x_1}, \quad y_2 = x_1 - 2$

2. Intersection of $y = 2\sqrt{x}$ and $x + y = 2$: Substitute $y = 2 - x$ into $y = 2\sqrt{x}$: $2 - x = 2\sqrt{x}$ Square both sides: $(2 - x)^2 = 4x$ Expand and simplify: $4 - 4x + x^2 = 4x$ $x^2 - 8x + 4 = 0$ This is the same quadratic equation as before, so the solutions are the same.

3. Intersection of $y = x - 2$ and $x + y = 2$: Substitute $y = x - 2$ into $x + y = 2$: $x + (x - 2) = 2$ $2x - 2 = 2$ $2x = 4 \quad \Rightarrow \quad x = 2$ Substituting back into $y = x - 2$, we get $y = 0$.

   Therefore, the intersection point is $(2, 0)$.

Step 2: Set up the integral

The area of the region can be found by integrating with respect to $x$. We need to compute the area under the curves $y = 2\sqrt{x}$ and $y = x - 2$, between the points of intersection.

The limits of integration are $x = 0$ and $x = 2$.

$\text{Area} = \int_{0}^{2} \left[ (2 - x) - (x - 2) \right] dx$

Step 3: Evaluate the integral

Simplify the integrand: $\text{Area} = \int_{0}^{2} \left(4 - 2x \right) dx$ Now, integrate: $\text{Area} = \left[4x - x^2 \right]_{0}^{2}$

Substitute the limits: [ \text{Area} = \left[4(2) - (2)^2 \right] - \left[4(0) - (0)^2 \right] ] [ \text{Area} = \left[8 - 4\right] - \left[0\right] ] $\text{Area} = 4 \text{ square units}$

Final Answer

The area of the region enclosed by the given curves is 4 square units.

Would you like to go through any of the steps in more detail or explore related questions?

Here are some related questions:

1. What are other methods for solving definite integrals when curves are involved?
2. How do you handle cases where the integration bounds aren't straightforward?
3. How does symmetry in the region simplify the calculation of area?
4. Can substitution methods simplify the integration of more complex curves?
5. How can software tools assist in solving similar problems?

Tip: Always double-check the integration limits and the order of subtraction when setting up the area integral to avoid errors.