$$\text { Given, } f(x)=\int_\limits x^{x^2+\frac{\pi}{6}} 2 \cos ^2 t d t \forall x \in \mathrm{R}$$

$$\text { As we know, if } \mathrm{I}(x)=\int_\limits{g(x)}^{h(x)} \phi(t) d t \text {, then }$$

$$\mathrm{I}^{\prime}(x)=\phi\{h(x)\} h^{\prime}(x)-\phi\{g(x)\} g^{\prime}(x)$$

$$\Rightarrow f(x)=2\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2 \cdot \frac{d}{d x}\left(x^2+\frac{\pi}{6}\right) -2 \cos ^2 x \cdot \frac{d x}{d x}$$

$$\Rightarrow f(x)=4 x\left\{\cos \left(x^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 x$$

Putting $$x=a$$ in the above equation, we get

$$f(a)=4 a\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 a$$

Also, the area of the region bounded by $$x=0$$,

$$\begin{aligned} & y=0, y=f(x) \text { and } x=a \text { is } \int_0^a f(x) d x \\ & \Rightarrow f(a)+2=\int_\limits0^a f(x) d x \\ & \Rightarrow 4 a\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2-2 \cos ^2 a+2=\int_\limits0^a f(x) d x \end{aligned}$$

Differentiating above equation w.r.t. a, we get

$$\begin{aligned} \Rightarrow & -4 a \cdot 2 \cos \left(a^2+\frac{\pi}{6}\right) \cdot \sin \left(a^2+\frac{\pi}{6}\right) \\ & 2 a+4\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2 \\ & -4 \cos a(-\sin a)=f(a) \\ \Rightarrow & -8 a^2 \sin \left(2 a^2+\frac{\pi}{3}\right)+4\left\{\cos \left(a^2+\frac{\pi}{6}\right)\right\}^2 \\ & +2 \sin 2 a=f(a)\{\because 2 \sin x \cos x=\sin 2 x\} \end{aligned}$$

Putting $$a=0$$ in the above equation, we get.

$$\begin{aligned} & 0+4 \cos ^2\left(\frac{\pi}{6}\right)+2 \sin (0)=f(0) \\ \Rightarrow & f(0)=4\left(\frac{\sqrt{3}}{2}\right)^2 \quad\left\{\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right\} . \\ \Rightarrow & f(0)=3 \end{aligned}$$

(i) Use if $$I(x)=\int_\limits{g(x)}^{h(x)} \phi(t) d t$$, then $$\mathrm{I}^{\prime}(x)=\phi\{h(x)\} h^{\prime}(x)-\phi\left\{g(x) g^{\prime}(x)\right.$$

(ii) Use the area of the region bounded by $$x=0, y=0, y=g(x)$$ and $$x=k$$ is $$\int_\limits0^k g(x) d x$$

(iii) Use the product rule of differentiation for further simplification.