## Step : Find the of Intersection have three curves- \(y = \sqrt{x}\) - \(y = \frac{3}{2} - \frac{x}{2}\) - \(y = \) (the \(x\)-axis) Let's find the intersection points in the first quadrant. ### **Intersection of \(y = \sqrt{x}\) and \(y = \frac{3}{2} - \frac{x}{2}\):** Set: \[ \sqrt{x} = \frac{3}{2} - \frac{x}{2} \] Let \(u = \sqrt{x}\), so \(x = u^2\): \[ u = \frac{3}{2} - \frac{u^2}{2} \] \[ 2u = 3 - u^2 \] \[ u^2 + 2u - 3 = \] \[ (u + 3)(u - 1) = \Rightarrow u = -3 \text{ or } u = 1 \] But \(u = \sqrt{x} \geq \), so \(u = 1\), \(x = 1\). So, intersection point: \((1, 1)\). ### **Intersection of \(y = \sqrt{x}\) and \(y = \):** \[ \sqrt{x} = \implies x = \] ### **Intersection of \(y = \frac{3}{2} - \frac{x}{2}\) and \(y = \):** \[ \frac{3}{2} - \frac{x}{2} = \implies x = 3 \] **Explanation:** Now we know the region is bounded from \(x = \) to \(x = 1\) by \(y = \sqrt{x}\) and from \(x = 1\) to \(x = 3\) by \(y = \frac{3}{2} - \frac{x}{2}\), both above the \(x\)-axis. --- ## Step 2: Set Up and Evaluate the Area We break the area into two parts: \[ \text{Area} = \int_{}^{1} \sqrt{x} \, dx + \int_{1}^{3} \left( \frac{3}{2} - \frac{x}{2} \right) dx \] ### **First Integral:** \[ \int_{}^{1} \sqrt{x} \, dx = \int_{}^{1} x^{1/2} dx = \left[ \frac{2}{3} x^{3/2} \right]_{}^{1} = \frac{2}{3} \] ### **Second Integral:** \[ \int_{1}^{3} \left( \frac{3}{2} - \frac{x}{2} \right) dx = \left[ \frac{3}{2}x - \frac{1}{4}x^2 \right]_{1}^{3} \] Evaluate at \(x = 3\): \[ \frac{3}{2} \times 3 - \frac{1}{4} \times 9 = \frac{9}{2} - \frac{9}{4} = \frac{9}{2} - \frac{9}{4} = \frac{18-9}{4} = \frac{9}{4} \] At \(x = 1\): \[ \frac{3}{2} \times 1 - \frac{1}{4} \times 1 = \frac{3}{2} - \frac{1}{4} = \frac{6-1}{4} = \frac{5}{4} \] So, the second area: \[ \frac{9}{4} - \frac{5}{4} = 1 \] **Explanation:** We split the area based on which curve is on top in each interval and integrated accordingly. --- ## **Final Answer** \[ \boxed{\frac{2}{3} + 1 = \frac{5}{3}} \] **The area of the region in the first quadrant is \(\frac{5}{3}\) square units.**