A simple model for the flow of air in and out of the lungs of a certain mammal is given by the following equation, where \( V(t) \) (measured in liters) is the volume of air in the lungs at time \( t \geq 0 \), \( t \) is measured in seconds, and \( t=0 \) corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs is exchanged with each breath. The amount that is exchanged is called the tidal volume. Complete parts a through \( c \) below. \( V^{\prime}(t)=-\frac{\pi}{6} \sin \left(\frac{\pi t}{6}\right) \) a. Find the volume function \( V \), assuming that \( V(0)=6 \) L. Notice that \( V \) changes over time at a known rate, \( V^{\prime} \). Which equation below correctly gives the volume function? A. \( V(0)=V(t)+\int_{0}^{t} V^{\prime}(x) d x \). C. \( V(t)=V(0)+\int_{a}^{b} V^{\prime}(t) d t \). \( V(t)=V(0)+\int_{0}^{t} V^{\prime}(x) d x \).

To find the volume function \( V(t) \) given the differential equation and the initial condition, let's follow the steps below. ### Given: \[ V'(t) = -\frac{\pi}{6} \sin\left(\frac{\pi t}{6}\right) \] \[ V(0) = 6 \text{ liters} \] ### Step 1: Integrate \( V'(t) \) to Find \( V(t) \) To find \( V(t) \), we integrate \( V'(t) \) with respect to \( t \): \[ V(t) = V(0) + \int_{0}^{t} V'(x) \, dx \] Substituting \( V'(x) \): \[ V(t) = 6 + \int_{0}^{t} \left(-\frac{\pi}{6} \sin\left(\frac{\pi x}{6}\right)\right) dx \] ### Step 2: Evaluate the Integral Let's evaluate the integral: \[ \int \left(-\frac{\pi}{6} \sin\left(\frac{\pi x}{6}\right)\right) dx \] Using the substitution \( u = \frac{\pi x}{6} \), \( du = \frac{\pi}{6} dx \), we get: \[ -\frac{\pi}{6} \int \sin(u) \cdot \frac{6}{\pi} du = -\int \sin(u) \, du = \cos(u) + C \] Substituting back \( u = \frac{\pi x}{6} \): \[ \cos\left(\frac{\pi x}{6}\right) + C \] Evaluating the definite integral from 0 to \( t \): \[ \left[\cos\left(\frac{\pi t}{6}\right) - \cos(0)\right] = \cos\left(\frac{\pi t}{6}\right) - 1 \] ### Step 3: Combine the Results Substituting the evaluated integral back into the expression for \( V(t) \): \[ V(t) = 6 + \left(\cos\left(\frac{\pi t}{6}\right) - 1\right) = 5 + \cos\left(\frac{\pi t}{6}\right) \] ### Conclusion The volume function \( V(t) \) is correctly given by: \[ V(t) = V(0) + \int_{0}^{t} V'(x) \, dx \] **Answer:** Option C. \( V(t)=V(0)+\int_{0}^{t} V^{\prime}(x) d x \).