Assertion (A) : The area bounded by one of the arcs of `y=cos` ax and X-axis is 2/a s.units.
Reason ( R ): The area bounded by `y=f(x)gt0` and y = 0 between x = a and x = b is `underset(a)overset(b)int` ydx The correct answer is

Options

A. Both (A) and ( R ) are true and R is the correct explanation of A
B. Both (A) and ( R ) are true and R is not the correct explanation of A
C. (A) is true, ( R ) is false
D. (A) false ( R ) is true

Answer

To solve the problem, we need to analyze the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion (A)**:
The assertion states that the area bounded by one of the arcs of \( y = \cos(ax) \) and the x-axis is \( \frac{2}{a} \) square units.

2. **Finding the Area Under the Curve**:
To find the area under the curve \( y = \cos(ax) \), we need to determine the limits for one complete arc of the cosine function. The cosine function completes one full cycle from \( 0 \) to \( 2\pi \). However, since we are looking for the area bounded by the x-axis, we will consider the interval where the cosine function is non-negative.

For \( y = \cos(ax) \):
- The function is non-negative between \( x = -\frac{\pi}{2a} \) and \( x = \frac{\pi}{2a} \) for one complete arc.

3. **Setting Up the Integral**:
The area \( A \) can be calculated using the integral:
\[ A = \int_{-\frac{\pi}{2a}}^{\frac{\pi}{2a}} \cos(ax) \, dx \]

4. **Evaluating the Integral**:
To evaluate the integral, we use the substitution \( u = ax \), which gives \( du = a \, dx \) or \( dx = \frac{du}{a} \). The limits change accordingly:
- When \( x = -\frac{\pi}{2a} \), \( u = -\frac{\pi}{2} \)
- When \( x = \frac{\pi}{2a} \), \( u = \frac{\pi}{2} \)

Thus, the integral becomes:
\[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(u) \cdot \frac{1}{a} \, du \]
\[ A = \frac{1}{a} \left[ \sin(u) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{a} \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right) \]
\[ A = \frac{1}{a} (1 - (-1)) = \frac{2}{a} \]

5. **Conclusion**:
We find that the area bounded by one arc of \( y = \cos(ax) \) and the x-axis is indeed \( \frac{2}{a} \) square units, confirming the assertion.

6. **Understanding the Reason (R)**:
The reason states that the area bounded by \( y = f(x) \) (where \( f(x) > 0 \)) and the x-axis between \( x = a \) and \( x = b \) is given by the integral \( \int_a^b f(x) \, dx \). This is a standard result in calculus and is indeed true.

### Final Answer:
Both the assertion (A) and the reason (R) are correct, and the reason provides a correct explanation for the assertion.