Make a rough sketch of the graph of `y = cos ^(2) x, 0 le x le (pi)/(2)` and find the area enclosed between the curve and the axes.

Answer

To solve the problem of finding the area enclosed between the curve \( y = \cos^2 x \) and the axes for \( 0 \leq x \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Sketch the Graph
1. **Identify the function**: We are given \( y = \cos^2 x \).
2. **Evaluate the function at the endpoints**:
- At \( x = 0 \): \( y = \cos^2(0) = 1^2 = 1 \)
- At \( x = \frac{\pi}{2} \): \( y = \cos^2\left(\frac{\pi}{2}\right) = 0^2 = 0 \)
3. **Behavior of the function**: The function \( \cos^2 x \) is continuous and decreases from 1 to 0 as \( x \) goes from 0 to \( \frac{\pi}{2} \).
4. **Rough sketch**: Draw the x-axis and y-axis. Plot the points (0, 1) and \(\left(\frac{\pi}{2}, 0\right)\). The curve will start at (0, 1) and smoothly decrease to \(\left(\frac{\pi}{2}, 0\right)\).

### Step 2: Set Up the Integral
To find the area \( A \) enclosed between the curve and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \), we set up the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \cos^2 x \, dx \]

### Step 3: Use the Trigonometric Identity
We can simplify \( \cos^2 x \) using the identity:
\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]
Thus, we rewrite the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} \, dx \]

### Step 4: Simplify the Integral
Now, we can separate the integral:
\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 + \cos(2x)) \, dx = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, dx + \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \right) \]

### Step 5: Calculate Each Integral
1. **Calculate \( \int_{0}^{\frac{\pi}{2}} 1 \, dx \)**:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

2. **Calculate \( \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \)**:
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]
Evaluating from 0 to \( \frac{\pi}{2} \):
\[ \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} (\sin(\pi) - \sin(0)) = \frac{1}{2} (0 - 0) = 0 \]

### Step 6: Combine the Results
Now substituting back into our area calculation:
\[ A = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

### Final Answer
The area enclosed between the curve \( y = \cos^2 x \) and the axes from \( 0 \) to \( \frac{\pi}{2} \) is:
\[ \boxed{\frac{\pi}{4}} \text{ square units} \]