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. $$ V(t)=V(0)+\int_{a}^{b} V^{\prime}(t) d t $$. $$ V(t)=V(0)+\int_{0}^{t} V^{\prime}(x) d x $$.

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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. $$ V(t)=V(0)+\int_{0}^{t} V^{\prime}(x) d x $$.
The volume function $$ V(t) $$ is given by: $$ V(t) = 5 + \cos\left(\frac{\pi t}{6}\right) $$ So, the correct option is C: $$ V(t) = V(0) + \int_{0}^{t} V'(x) \, dx $$.

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