Let $f:[0,2] \in R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$ for $x \in[0,2]$. If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$, then $F(2)$ equals

$e^4-1$

We have, $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$

∴   $F'(x)=2 x f(x)$

$\Rightarrow f'(x)=2 x f(x)$            $\left[∵ F'(x)=f'(x)\right]$

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

$\Rightarrow \log (f(x))=x^2+\log C$

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

$\Rightarrow f(0)=C e^0=C$

$\Rightarrow 1=C$                 [∵  f(0) = 1]

∴   $f(x)=e^{x^2}$

Now, $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$ and $f(x)=e^{x^2}$

$\Rightarrow F(x)=\int\limits_0^{x^2} e^t d t=e^{x^2}-1$

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