1. **Problem Statement:** Calculate the definite integrals of the function $f(x)$ whose graph consists of semicircles below and above the x-axis on given intervals. 2. **Understanding the graph:** - From $x=0$ to $x=2$, $f(x)$ is a semicircle below the x-axis. - From $x=2$ to $x=6$, $f(x)$ is a semicircle above the x-axis. - The curve intersects the x-axis at $x=0, 2, 4, 6$. 3. **Formula for area of a semicircle:** $$\text{Area} = \frac{1}{2} \pi r^2$$ where $r$ is the radius of the semicircle. 4. **Calculate radii:** - For $0 \leq x \leq 2$, radius $r=2$ (since the semicircle spans 2 units). - For $2 \leq x \leq 4$, radius $r=2$. - For $4 \leq x \leq 6$, radius $r=2$. 5. **Calculate each integral:** A. $\int_0^2 f(x) \, dx$ is the area of the semicircle below x-axis, so integral is negative: $$\int_0^2 f(x) \, dx = -\frac{1}{2} \pi (2)^2 = -2\pi \approx -6.28$$ B. $\int_1^4 f(x) \, dx$ covers part of the semicircle below x-axis from 1 to 2 and the semicircle above x-axis from 2 to 4. - Area from 1 to 2 (half of semicircle below x-axis): half of $-2\pi$ is $-\pi \approx -3.14$ - Area from 2 to 4 (full semicircle above x-axis): $+2\pi \approx +6.28$ - Total: $$-3.14 + 6.28 = 3.14$$ C. $\int_1^4 |f(x)| \, dx$ is the sum of absolute areas: $$3.14 + 6.28 = 9.42$$ D. $\int_4^2 f(x) \, dx = -\int_2^4 f(x) \, dx = -2\pi \approx -6.28$$ E. $\int_4^4 f(x) \, dx = 0$ (integral over zero width interval). **Final answers rounded to 2 decimals:** A. $-6.28$ B. $3.14$ C. $9.42$ D. $-6.28$ E. $0.00$