The problem asks for the area of a region in the first quadrant bounded by the curve $y = \sqrt{x}$, the line $y = 2 - x$, and the $x$-axis ($y = 0$).

Finding Boundaries and Intersection Points

• Curve 1: $y = \sqrt{x}$ (starts at (0,0), goes upwards)
• Curve 2: $y = 2 - x$ (a line crossing the y-axis at 2 and x-axis at 2)
• Curve 3: $y = 0$ (the x-axis)
• Intersection of $y = \sqrt{x}$ and $y = 2 - x$: Set $\sqrt{x} = 2 - x$. Squaring both sides gives $x = (2 - x)^2 = 4 - 4x + x^2$. Rearranging, we get $x^2 - 5x + 4 = 0$, which factors into $(x-1)(x-4) = 0$. The possible solutions are $x=1$ and $x=4$. If $x=1$, $y = \sqrt{1} = 1$ and $y = 2 - 1 = 1$. So, (1, 1) is a valid intersection point. If $x=4$, $y = \sqrt{4} = 2$ and $y = 2 - 4 = -2$. This point (4, -2) is not in the first quadrant and $y=\sqrt{x}$ implies $y \ge 0$. Thus, $x=4$ is an extraneous solution.
• The region lies in the first quadrant. The intersection point relevant to the bounded area is (1, 1). The boundaries intersect the x-axis at $x=0$ (for $y=\sqrt{x}$) and $x=2$ (for $y=2-x$).

Calculating the Area using Integration

The region needs to be split into two parts for integration with respect to $x$, divided at the intersection point $x=1$.

• Part 1: From $x=0$ to $x=1$. The region is bounded above by $y = \sqrt{x}$ and below by the $x$-axis ($y=0$). The area $A_1$ is calculated as:

  $A_1 = \int_{0}^{1} (\sqrt{x} - 0) \, dx = \int_{0}^{1} x^{1/2} \, dx$

  $A_1 = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1}$

  $A_1 = \frac{2}{3} (1)^{3/2} - \frac{2}{3} (0)^{3/2} = \frac{2}{3}$

• Part 2: From $x=1$ to $x=2$. The region is bounded above by $y = 2 - x$ and below by the $x$-axis ($y=0$). The area $A_2$ is calculated as:

  $A_2 = \int_{1}^{2} ((2 - x) - 0) \, dx = \int_{1}^{2} (2 - x) \, dx$

  $A_2 = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2}$

  $A_2 = \left( 2(2) - \frac{2^2}{2} \right) - \left( 2(1) - \frac{1^2}{2} \right)$

  $A_2 = (4 - 2) - (2 - \frac{1}{2}) = 2 - \frac{3}{2} = \frac{1}{2}$

• Total Area: The total area $A$ is the sum of $A_1$ and $A_2$.

  $A = A_1 + A_2 = \frac{2}{3} + \frac{1}{2}$

  $A = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}$

Conclusion

The area of the specified region in the first quadrant is $\frac{7}{6}$.