1. **State the problem:** Calculate the value of $$\sqrt{\frac{\int_{3\pi/8}^{7\pi/8} 40 \, dx + \int_{7\pi/8}^{2\pi + 3\pi/8} (-80) \, dx}{2\pi}}$$ 2. **Recall the integral formula:** For a constant $c$, the definite integral over $[a,b]$ is $$\int_a^b c \, dx = c(b - a)$$ 3. **Calculate each integral:** - First integral: $$\int_{3\pi/8}^{7\pi/8} 40 \, dx = 40 \times \left(\frac{7\pi}{8} - \frac{3\pi}{8}\right) = 40 \times \frac{4\pi}{8} = 40 \times \frac{\pi}{2} = 20\pi$$ - Second integral: $$\int_{7\pi/8}^{2\pi + 3\pi/8} (-80) \, dx = -80 \times \left(2\pi + \frac{3\pi}{8} - \frac{7\pi}{8}\right) = -80 \times \left(2\pi - \frac{4\pi}{8}\right) = -80 \times \left(2\pi - \frac{\pi}{2}\right) = -80 \times \frac{3\pi}{2} = -120\pi$$ 4. **Sum the integrals:** $$20\pi + (-120\pi) = -100\pi$$ 5. **Divide by $2\pi$:** $$\frac{-100\pi}{2\pi} = -50$$ 6. **Take the square root:** Since the value inside the square root is negative ($-50$), the expression $$\sqrt{-50}$$ is not a real number. It is an imaginary number: $$\sqrt{-50} = \sqrt{50}i = 5\sqrt{2}i$$ **Final answer:** $$5\sqrt{2}i$$ This means the original expression evaluates to an imaginary number $5\sqrt{2}i$ because the value inside the square root is negative.