The area bounded by `f(x)=sin^2x` and the x - axis from x = a to x = b , where `f''(a)=f''(b)=0(AAa,b,in(0,pi))` is

Options

A. `pi/4`
B. `pi+2`
C. 2
D. `(pi+2)/(4)`

Answer

To find the area bounded by the curve \( f(x) = \sin^2 x \) and the x-axis from \( x = a \) to \( x = b \), where \( f''(a) = f''(b) = 0 \) and \( a, b \in (0, \pi) \), we can follow these steps:

### Step 1: Find the second derivative of \( f(x) \)
First, we need to calculate the first and second derivatives of \( f(x) \).

1. **First derivative**:
\[ f'(x) = 2 \sin x \cos x = \sin(2x) \]

2. **Second derivative**:
\[ f''(x) = 2 \cos(2x) \]

### Step 2: Set the second derivative to zero
We need to find the points where \( f''(x) = 0 \):
\[ 2 \cos(2x) = 0 \implies \cos(2x) = 0 \]

### Step 3: Solve for \( x \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ 2x = (2n + 1) \frac{\pi}{2} \implies x = \frac{(2n + 1) \pi}{4} \]

For \( n = 0 \):
\[ x = \frac{\pi}{4} \]

For \( n = 1 \):
\[ x = \frac{3\pi}{4} \]

### Step 4: Identify the values of \( a \) and \( b \)
Since \( a \) and \( b \) must lie in the interval \( (0, \pi) \), we have:
\[ a = \frac{\pi}{4}, \quad b = \frac{3\pi}{4} \]

### Step 5: Set up the integral for the area
The area \( A \) bounded by the curve and the x-axis from \( x = a \) to \( x = b \) is given by:
\[ A = \int_{a}^{b} f(x) \, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin^2 x \, dx \]

### Step 6: Use the identity for \( \sin^2 x \)
We can use the identity:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Thus, the integral becomes:
\[ A = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1 - \cos(2x)}{2} \right) dx \]

### Step 7: Break the integral
This can be split into two parts:
\[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 1 \, dx - \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos(2x) \, dx \]

### Step 8: Evaluate the integrals
1. **First integral**:
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 1 \, dx = \left[ x \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2} \]

2. **Second integral**:
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = \frac{1}{2} \left( \sin\left(\frac{3\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) \right) = \frac{1}{2} \left( -1 - 1 \right) = -1 \]

### Step 9: Combine the results
Putting it all together:
\[ A = \frac{1}{2} \left( \frac{\pi}{2} \right) - \frac{1}{2} (-1) = \frac{\pi}{4} + \frac{1}{2} \]

### Final Result
Thus, the area bounded by the curve \( f(x) = \sin^2 x \) and the x-axis from \( x = a \) to \( x = b \) is:
\[ A = \frac{\pi + 2}{4} \]