Given, $$\mathrm{F}^{\prime}(x)=f^{\prime}(x)\quad \text{.... (i)}$$

And $$\quad \mathrm{F}(x)=\int_0^{x^2} f(\sqrt{t}) d t$$

Differentiate the equation (ii) w.r.t. $$x$$

$$\begin{aligned} & \Rightarrow \quad \mathrm{F}^{\prime}(x)=\frac{d\left(x^2\right)}{d x} \cdot f\left(\sqrt{x^2}\right)-\frac{d(0)}{d x} \cdot f(\sqrt{0}) \\ & \Rightarrow \quad f^{\prime}(x)=2 x \cdot f(x)-0 \\ & \Rightarrow \quad \frac{d f(x)}{f(x)}=2 x d x \\ & \Rightarrow \int \frac{d f(x)}{f(x)}=\int 2 x d x \\ & \Rightarrow \ln f(x)=2 \cdot \frac{x^2}{2}+\mathrm{C} \\ & \Rightarrow \ln f(x)=x^2+\mathrm{C} \quad \text{... (iii)} \end{aligned}$$

Given, $$f(0)=1$$

$$\begin{array}{llrl} & \therefore & \ln (1) & =0^2+C \\ & \Rightarrow & 0 & =0+C \\ \Rightarrow & & C=0 \end{array}$$

Put $$c=0$$ in the equation (iii)

$$\begin{array}{ll} \Rightarrow & \ln f(x)=x^2 \\ \Rightarrow \quad f(x) & =e^{x^2} \\ \Rightarrow \quad f(\sqrt{t}) & =e^t \end{array}$$

Put $$f(\sqrt{t})=e^t$$ in the equation (ii)

$$\begin{array}{ll} \Rightarrow & \mathrm{F}(x)=\int_0^{x^2} e^t \cdot d t \\ \Rightarrow & \mathrm{F}(x)=\left[e^t\right]_0^{x^2} \\ \Rightarrow & \mathrm{F}(x)=e^{x^2}-1 \end{array}$$

Put $$\quad x=2$$

$$\Rightarrow \mathrm{F}(2)=e^4-1$$

Hint:

(i) If $$y=\int_{\phi(x)}^{\psi(x)} f(t) d t$$, then $$\frac{d y}{d x}=\frac{d \psi(x)}{d x} \cdot f(\psi(x))-\frac{d \phi(x)}{d x} \cdot f(\phi(x))$$

(ii) $$\int \frac{f^{\prime}(x) d x}{f(x)}=\ln (f(x))+c$$