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 \( \mathbf{c} \) below. \[ V^{\prime}(t)=-\frac{\pi}{6} \sin \left(\frac{\pi t}{6}\right) \] a. Find the volume function V , assuming that \( \mathrm{V}(0)=6 \mathrm{~L} \). Notice that V changes over time at a known rate, \( \mathrm{V}^{\prime} \). Which equation below correctly gives the volume function? A. \( V(0)=V(t)+\int_{0}^{t} V^{\prime}(x) d x \) c. \( V(t)=V(0)+\int_{a}^{b} v^{\prime}(t) d t \). D. \( V(0)=V(t)+\int_{a}^{b} V^{\prime}(t) d t \). Find the volume function V , assuming that \( \mathrm{V}(0)=6 \mathrm{~L} \). \[ V(t)= \] \( \square \) (Type an exact answer.)

To find the volume function \( V(t) \) given the initial condition \( V(0) = 6 \) liters and the rate of change \( V'(t) = -\frac{\pi}{6} \sin\left(\frac{\pi t}{6}\right) \), follow these steps: 1. **Integrate the Rate of Change:** \[ V(t) = V(0) + \int_{0}^{t} V'(x) \, dx \] Substituting \( V'(x) \): \[ V(t) = 6 + \int_{0}^{t} \left(-\frac{\pi}{6} \sin\left(\frac{\pi x}{6}\right)\right) dx \] 2. **Compute the Integral:** Let \( u = \frac{\pi x}{6} \), hence \( du = \frac{\pi}{6} dx \) or \( dx = \frac{6}{\pi} du \). \[ \int \left(-\frac{\pi}{6} \sin(u)\right) \cdot \frac{6}{\pi} du = -\int \sin(u) \, du = \cos(u) + C \] Substituting back \( u = \frac{\pi x}{6} \): \[ \cos\left(\frac{\pi t}{6}\right) - \cos(0) = \cos\left(\frac{\pi t}{6}\right) - 1 \] 3. **Combine the Results:** \[ V(t) = 6 + \left(\cos\left(\frac{\pi t}{6}\right) - 1\right) = 5 + \cos\left(\frac{\pi t}{6}\right) \] **Final Volume Function:** \[ V(t) = 5 + \cos\left(\frac{\pi t}{6}\right) \] **Answer:** For all \( t \geq 0 \), the volume is \( V(t) = 5 + \cos\left(\dfrac{\pi t}{6}\right) \). \[ V(t) = 5 + \cos\left(\frac{\pi t}{6}\right) \]