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 function \( f(x) = \sin^2 x \) 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: Understanding the Function and Its Derivatives
The function given is \( f(x) = \sin^2 x \). We need to find the second derivative \( f''(x) \) and set it to zero to find the points \( a \) and \( b \).

### Step 2: Calculate the First Derivative
To find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^2 x) = 2\sin x \cos x = \sin(2x) \]

### Step 3: Calculate the Second Derivative
Next, we find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \]

### Step 4: Set the Second Derivative to Zero
We need to find the values of \( a \) and \( b \) such that:
\[ f''(a) = 2\cos(2a) = 0 \quad \text{and} \quad f''(b) = 2\cos(2b) = 0 \]
This implies:
\[ \cos(2a) = 0 \quad \text{and} \quad \cos(2b) = 0 \]

### Step 5: Solve for \( a \) and \( b \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ 2a = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad a = \frac{\pi}{4} + \frac{n\pi}{2} \]
\[ 2b = \frac{\pi}{2} + m\pi \quad \Rightarrow \quad b = \frac{\pi}{4} + \frac{m\pi}{2} \]
For \( a \) and \( b \) to be in the interval \( (0, \pi) \), we can take:
- \( a = \frac{\pi}{4} \)
- \( b = \frac{3\pi}{4} \)

### Step 6: Set Up the Integral for Area
The area \( A \) bounded by \( f(x) \) and the x-axis from \( a \) to \( b \) is given by:
\[ A = \int_a^b f(x) \, dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin^2 x \, dx \]

### Step 7: Use the Identity for Integration
Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
\[ A = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx \]
\[ = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (1 - \cos(2x)) \, dx \]

### Step 8: Evaluate the Integral
Now we evaluate the integral:
\[ = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \]
Calculating the limits:
\[ = \frac{1}{2} \left[ \left( \frac{3\pi}{4} - \frac{\sin(\frac{3\pi}{2})}{2} \right) - \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} \right) \right] \]
\[ = \frac{1}{2} \left[ \frac{3\pi}{4} + \frac{1}{2} - \frac{\pi}{4} + \frac{1}{2} \right] \]
\[ = \frac{1}{2} \left[ \frac{2\pi}{4} + 1 \right] = \frac{1}{2} \left[ \frac{\pi}{2} + 1 \right] \]

### Final Area Calculation
Thus, the area \( A \) is:
\[ A = \frac{\pi}{4} + \frac{1}{2} \]