Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f(0) = 1$. Let $F(x) = \int_{0}^{x^{2}} f(\sqrt{t})\,dt$ for $x \in [0, 2]$. If $F'(x) = f'(x)$ for all $x \in (0, 2)$, then $F(2)$ equals

$f \text{ continuous on } [0,2],\ \text{differentiable on } (0,2),\ f(0) = 1$

$F(x) = \int_0^{x^2} f(\sqrt{t})\,dt,\qquad F'(x) = f'(x)\ \ \forall\, x\in(0,2)$

$\text{To find: } F(2)$

$G(u) = \int_0^u f(\sqrt{t})\,dt \Rightarrow G'(u) = f(\sqrt{u})$

$F(x) = G(x^2) \Rightarrow F'(x) = G'(x^2)\cdot 2x = 2x\,f(x)$

$2x\,f(x) = f'(x) \Rightarrow \dfrac{f'(x)}{f(x)} = 2x$

$\ln|f(x)| = x^2 + C$

$f(0)=1 \Rightarrow C = 0 \Rightarrow f(x) = e^{x^2}$

$f(\sqrt{t}) = e^{t} \Rightarrow F(x) = \int_0^{x^2} e^t\,dt = e^{x^2} - 1$

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

$\boxed{F(2) = e^4 - 1 \ \text{— option } (2)}$