Let F(x) = int_x^{x^2+frac{pi}{6}} 2 cos^2 t | vedclass

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(B) Given $F'(a) + 2 = \int_0^a f(x) \, dx$.Differentiating both sides with respect to $a$ using the Fundamental Theorem of Calculus:$F''(a) = f(a)$.N

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$3$

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(B) Given $F'(a) + 2 = \int_0^a f(x) \, dx$.Differentiating both sides with respect to $a$ using the Fundamental Theorem of Calculus:$F''(a) = f(a)$.Now,we find $F'(x)$ using the Leibniz rule for differentiation under the integral sign:$F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$.$F'(x) = 2 \cos^2(x^2 + \frac{\pi}{6}) \cdot \frac{d}{dx}(x^2 + \frac{\pi}{6}) - 2 \cos^2(x) \cdot \frac{d}{dx}(x)$.$F'(x) = 2 \cos^2(x^2 + \frac{\pi}{6}) \cdot (2x) - 2 \cos^2(x)$.To find $f(0)$,we need $F''(0)$.$F''(x) = \frac{d}{dx} [4x \cos^2(x^2 + \frac{\pi}{6}) - 2 \cos^2(x)]$.$F''(x) = 4 \cos^2(x^2 + \frac{\pi}{6}) + 4x \cdot 2 \cos(x^2 + \frac{\pi}{6}) \cdot (-\sin(x^2 + \frac{\pi}{6})) \cdot 2x - 2 \cdot 2 \cos(x) \cdot (-\sin(x))$.$F''(x) = 4 \cos^2(x^2 + \frac{\pi}{6}) - 8x^2 \sin(2(x^2 + \frac{\pi}{6})) + 2 \sin(2x)$.Evaluating at $x=0$:$f(0) = F''(0) = 4 \cos^2(\frac{\pi}{6}) - 8(0)^2 \sin(2 \cdot \frac{\pi}{6}) + 2 \sin(0)$.$f(0) = 4 \cdot (\frac{\sqrt{3}}{2})^2 - 0 + 0$.$f(0) = 4 \cdot \frac{3}{4} = 3$.

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ 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

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