Let $F(x) =$ integral$_{x}^{x^{2} + \pi/6} 2 \cos^{2} t$ dt for all x in R and $f : [0, \frac{1}{2}] \to [0,$ inf) be a continuous function. For a in $[0, \frac{1}{2}],$ 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 ___

Let F(x)=∫ₓ^(x²+π/6) 2cos² t (dt) for all x ∈ R and f:[0,1/2]→[0,∞) be a continuous function. For a ∈ [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

$F(x) = \int_{x}^{x^{2}+\pi/6} 2\cos^{2}t\,dt$

$f:\left[0,\tfrac{1}{2}\right]\to[0,\infty) \text{ continuous}$

$F'(a)+2 = \text{area bounded by } x=0,\ y=0,\ y=f(x),\ x=a$

$\text{Find } f(0)$

$1.\ \text{find } F'(x) \text{ (Leibniz)}$

$2.\ \text{set area } = F'(a)+2$

$3.\ \text{differentiate } \Rightarrow f(0)$

$\text{area} = \int_0^a f(x)\,dx\ \ (f\ge 0)$

$F'(x) = 2\cos^{2}\!\left(x^{2}+\tfrac{\pi}{6}\right)\cdot 2x - 2\cos^{2}x\cdot 1$

$= 4x\cos^{2}\!\left(x^{2}+\tfrac{\pi}{6}\right) - 2\cos^{2}x$

$\int_0^a f(x)\,dx = F'(a)+2$

$\frac{d}{da}\int_0^a f(x)\,dx = f(a)$

$f(a) = F''(a) \Rightarrow f(0) = F''(0)$

$F''(x) = 4\cos^{2}\!\left(x^{2}+\tfrac{\pi}{6}\right) - 16x^{2}\cos\!\left(x^{2}+\tfrac{\pi}{6}\right)\sin\!\left(x^{2}+\tfrac{\pi}{6}\right) + 4\sin x\cos x$

$f(0) = F''(0) = 4\cos^{2}\!\left(\tfrac{\pi}{6}\right) = 4\cdot\frac{3}{4} = 3$

$\boxed{f(0) = 3}$