1. **State the problem:** Calculate the value of $$\sqrt{\frac{\int_{3\pi/8}^{7\pi/8} 40^2 \, dx}{2\pi} + \frac{\int_{7\pi/8}^{2\pi + 3\pi/8} (-80)^2 \, dx}{2\pi}}$$ 2. **Recall the integral of a constant:** For a constant $c$, $\int_a^b c^2 \, dx = c^2 (b - a)$. 3. **Calculate each integral:** - First integral: $$\int_{3\pi/8}^{7\pi/8} 40^2 \, dx = 1600 \times \left(\frac{7\pi}{8} - \frac{3\pi}{8}\right) = 1600 \times \frac{4\pi}{8} = 1600 \times \frac{\pi}{2} = 800\pi$$ - Second integral: $$\int_{7\pi/8}^{2\pi + 3\pi/8} (-80)^2 \, dx = 6400 \times \left(2\pi + \frac{3\pi}{8} - \frac{7\pi}{8}\right) = 6400 \times \left(2\pi - \frac{4\pi}{8}\right) = 6400 \times \left(2\pi - \frac{\pi}{2}\right) = 6400 \times \frac{3\pi}{2} = 9600\pi$$ 4. **Divide each integral by $2\pi$:** - First term: $$\frac{800\pi}{2\pi} = 400$$ - Second term: $$\frac{9600\pi}{2\pi} = 4800$$ 5. **Sum the two terms:** $$400 + 4800 = 5200$$ 6. **Take the square root:** $$\sqrt{5200} = \sqrt{100 \times 52} = 10 \sqrt{52} = 10 \times 2 \sqrt{13} = 20 \sqrt{13}$$ **Final answer:** $$20 \sqrt{13}$$