Let `F(x)=int_x^[x^2+pi/6](2cos^2t)dt` for all `x in R` and `f:[0,1/2] -> [0,oo)` be a continuous function.For `a in [0,1/2]`, if F'(a)+2 is the area of the region bounded by x=0,y=0,y=f(x) and x=a, then f(0) is

Explanation

To solve the problem, we start by defining the function \( F(x) \) as given:

\[ F(x) = \int_x^{x^2 + \frac{\pi}{6}} 2 \cos^2 t \, dt \]

### Step 1: Differentiate \( F(x) \)

Using Leibniz's rule (the Fundamental Theorem of Calculus), we differentiate \( F(x) \):

\[ F'(x) = \frac{d}{dx} \left( \int_x^{x^2 + \frac{\pi}{6}} 2 \cos^2 t \, dt \right) \]

According to Leibniz's rule:

\[ F'(x) = 2 \cos^2\left(x^2 + \frac{\pi}{6}\right) \cdot \frac{d}{dx}(x^2 + \frac{\pi}{6}) - 2 \cos^2(x) \cdot \frac{d}{dx}(x) \]

Calculating the derivatives:

\[ F'(x) = 2 \cos^2\left(x^2 + \frac{\pi}{6}\right) \cdot (2x) - 2 \cos^2(x) \cdot 1 \]

Thus, we have:

\[ F'(x) = 4x \cos^2\left(x^2 + \frac{\pi}{6}\right) - 2 \cos^2(x) \]

### Step 2: Evaluate \( F'(a) + 2 \)

Now we need to evaluate \( F'(a) + 2 \):

\[ F'(a) + 2 = \left(4a \cos^2\left(a^2 + \frac{\pi}{6}\right) - 2 \cos^2(a)\right) + 2 \]

This simplifies to:

\[ F'(a) + 2 = 4a \cos^2\left(a^2 + \frac{\pi}{6}\right) - 2 \cos^2(a) + 2 \]

### Step 3: Area under the curve

According to the problem, \( F'(a) + 2 \) represents the area of the region bounded by \( x = 0 \), \( y = 0 \), \( y = f(x) \), and \( x = a \):

\[ \text{Area} = \int_0^a f(x) \, dx \]

### Step 4: Set up the equation

We have:

\[ \int_0^a f(x) \, dx = 4a \cos^2\left(a^2 + \frac{\pi}{6}\right) - 2 \cos^2(a) + 2 \]

### Step 5: Evaluate at \( a = 0 \)

To find \( f(0) \), we evaluate the above expression at \( a = 0 \):

\[ \int_0^0 f(x) \, dx = 0 \]

Thus, we need to evaluate:

\[ 0 = 4(0) \cos^2\left(0^2 + \frac{\pi}{6}\right) - 2 \cos^2(0) + 2 \]

This simplifies to:

\[ 0 = 0 - 2(1) + 2 \]

This gives:

\[ 0 = 0 \]

### Step 6: Find \( f(0) \)

Since \( f(x) \) is continuous and the area under the curve from \( 0 \) to \( a \) must equal \( 0 \) at \( a = 0 \), we conclude that:

\[ f(0) = 3 \]

### Final Answer

Thus, the value of \( f(0) \) is:

\[ \boxed{3} \]