Let's break down the problem step by step: ### **Step 1: Understanding the Problem** We are asked to **evaluate the integral of the functions** given by the equations of three semicircles above the x-axis: - \( y = \sqrt{4 - (x-2)^2} \) - \( y = \sqrt{9 - (x-5)^2} \) - \( y = \sqrt{25 - (x-12)^2} \) These represent semicircles with different radii and centers. --- ### **Step 2: Find Center and Radius of Each Semicircle** The general equation for a circle is: \[ y = \sqrt{r^2 - (x-h)^2} \] where **center = (h, 0)** and **radius = r**. #### 1. \( y = \sqrt{4 - (x-2)^2} \) - Center: (2, 0) - Radius: \( \sqrt{4} = 2 \) #### 2. \( y = \sqrt{9 - (x-5)^2} \) - Center: (5, 0) - Radius: \( \sqrt{9} = 3 \) #### 3. \( y = \sqrt{25 - (x-12)^2} \) - Center: (12, 0) - Radius: \( \sqrt{25} = 5 \) --- ### **Step 3: Area Under Each Semicircle** The area under \( y = \sqrt{r^2 - (x-h)^2} \) from \( x = h-r \) to \( x = h+r \) is **half the area of a circle**, i.e., \[ \text{Area} = \frac{1}{2} \pi r^2 \] #### a) First Semicircle: \[ \text{Area}_1 = \frac{1}{2} \pi (2)^2 = 2\pi \] #### b) Second Semicircle: \[ \text{Area}_2 = \frac{1}{2} \pi (3)^2 = \frac{9}{2}\pi \] #### c) Third Semicircle: \[ \text{Area}_3 = \frac{1}{2} \pi (5)^2 = \frac{25}{2}\pi \] --- ### **Step 4: Add the Areas** \[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 \] \[ = 2\pi + \frac{9}{2}\pi + \frac{25}{2}\pi \] Combine the terms: \[ = 2\pi + \frac{34}{2}\pi = 2\pi + 17\pi = 19\pi \] --- ### **Final Answer** \[ \boxed{19\pi} \] This is the area under the curves, or equivalently, the value of the integrals of these functions over their respective domains, added together.