The area contained between the x-axis and one area of the curve `y=cos 3x,` is

Options

A. `1//3`
B. `2//3`
C. `2//7`
D. `2//5`

Answer

To find the area contained between the x-axis and one loop of the curve \( y = \cos(3x) \), we will follow these steps:

### Step 1: Determine the limits of integration
The function \( y = \cos(3x) \) completes one full cycle (one loop) between the points where it crosses the x-axis. To find these points, we set \( \cos(3x) = 0 \). This occurs when:
\[ 3x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For the first loop, we can take \( n = 0 \) and \( n = 1 \):
1. For \( n = 0 \): \( 3x = \frac{\pi}{2} \) → \( x = \frac{\pi}{6} \)
2. For \( n = 1 \): \( 3x = \frac{3\pi}{2} \) → \( x = \frac{\pi}{2} \)

Thus, the limits of integration for one loop of the curve are from \( x = -\frac{\pi}{6} \) to \( x = \frac{\pi}{6} \).

### Step 2: Set up the integral for the area
The area \( A \) between the curve and the x-axis can be calculated using the integral:
\[ A = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos(3x) \, dx \]

### Step 3: Calculate the integral
To solve the integral, we first find the antiderivative of \( \cos(3x) \):
\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C \]

Now, we can evaluate the definite integral:
\[ A = \left[ \frac{1}{3} \sin(3x) \right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \]

### Step 4: Evaluate the limits
Calculating at the upper limit \( x = \frac{\pi}{6} \):
\[ \sin\left(3 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
Calculating at the lower limit \( x = -\frac{\pi}{6} \):
\[ \sin\left(3 \cdot -\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{2}\right) = -1 \]

Now substituting these values into the integral:
\[ A = \frac{1}{3} \left(1 - (-1)\right) = \frac{1}{3} (1 + 1) = \frac{1}{3} \cdot 2 = \frac{2}{3} \]

### Final Answer
The area contained between the x-axis and one loop of the curve \( y = \cos(3x) \) is:
\[ \boxed{\frac{2}{3}} \text{ square units} \]

---