Determine Function Value from Leibniz Rule and Area Under Curve

To find the value of $f(0)$, we first express the given condition mathematically.

Given the function: $F(x) = \int_{x}^{x^2 + \frac{\pi}{6}} 2\cos^2 t \, dt$

The area of the region bounded by $x = 0$, $y = 0$, $y = f(x)$, and $x = a$ is given by the definite integral: $A(a) = \int_{0}^{a} f(x) \, dx$ since $f(x) \ge 0$ for all $x \in \left[0, \frac{1}{2}\right]$.

According to the given condition, for any $a \in \left[0, \frac{1}{2}\right]$: $F'(a) + 2 = \int_{0}^{a} f(x) \, dx$

First, we find $F'(x)$ using the Leibniz rule of differentiation under the integral sign: $F'(x) = 2\cos^2\left(x^2 + \frac{\pi}{6}\right) \cdot \frac{d}{dx}\left(x^2 + \frac{\pi}{6}\right) - 2\cos^2(x) \cdot \frac{d}{dx}(x)$ $F'(x) = 4x\cos^2\left(x^2 + \frac{\pi}{6}\right) - 2\cos^2(x)$

Notice that at $a = 0$: $F'(0) = 4(0)\cos^2\left(\frac{\pi}{6}\right) - 2\cos^2(0) = -2$ $F'(0) + 2 = -2 + 2 = 0 = \int_{0}^{0} f(x) \, dx$ which is consistent with the given condition.

Now, differentiating both sides of the equation $F'(a) + 2 = \int_{0}^{a} f(x) \, dx$ with respect to $a$, by the Fundamental Theorem of Calculus, we obtain: $F''(a) = f(a)$

Next, we differentiate $F'(x)$ with respect to $x$ to find $F''(x)$: $F''(x) = \frac{d}{dx}\left[ 4x\cos^2\left(x^2 + \frac{\pi}{6}\right) - 2\cos^2(x) \right]$ $F''(x) = 4\cos^2\left(x^2 + \frac{\pi}{6}\right) + 4x \cdot \frac{d}{dx}\left[\cos^2\left(x^2 + \frac{\pi}{6}\right)\right] - 2 \cdot 2\cos(x)(-\sin(x))$ $F''(x) = 4\cos^2\left(x^2 + \frac{\pi}{6}\right) - 16x^2\cos\left(x^2 + \frac{\pi}{6}\right)\sin\left(x^2 + \frac{\pi}{6}\right) + 2\sin(2x)$

Since $f$ is continuous, we evaluate $f(0) = F''(0)$: $f(0) = 4\cos^2\left(\frac{\pi}{6}\right) - 0 + 0$

Using $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$: $f(0) = 4 \left(\frac{\sqrt{3}}{2}\right)^2 = 4 \times \frac{3}{4} = 3$

Thus, the value of $f(0)$ is $3$.