1. Given $F(x)=\int_0^{x^2} f(\sqrt t)\,dt, \quad x\in[0,2]$ with $f(0)=1, \quad F'(x)=f'(x).$

2. Simplify $F(x)$

   Use the substitution $t=u^2 \implies dt=2u\,du.$ When $t=0$, $u=0$; when $t=x^2$, $u=x$ (since $x\in[0,2]$).

   Therefore, $F(x)=\int_0^x f(u)\,2u\,du=\int_0^x 2u f(u)\,du.$

3. Differentiate $F(x)$

   By the Fundamental Theorem of Calculus, $F'(x)=2x f(x).$

   But given that $F'(x)=f'(x),$ so for all $x\in[0,2]$, $f'(x)=2x f(x).$

4. Solve the differential equation

   $\frac{f'(x)}{f(x)}=2x.$ Integrating, $\ln f(x)=x^2+C.$ Hence, $f(x)=Ce^{x^2}.$

   Using $f(0)=1$, $1=Ce^0 \implies C=1.$ Thus, $f(x)=e^{x^2}.$

5. Find $F(2)$

   Since $F'(x)=f'(x)$, we have $F(x)=f(x)+K$ for some constant $K$.

   Now, $F(0)=\int_0^0 f(\sqrt t)\,dt=0,$ and $f(0)=1.$ So, $0=1+K \implies K=-1.$

   Therefore, $F(x)=f(x)-1=e^{x^2}-1.$

   Hence, $F(2)=e^4-1.$

6. Check options

   • A: $e^2-1$ ❌
   • B: $e^4-1$ ✅
   • C: $e-1$ ❌
   • D: $e^4$ ❌

Therefore, the correct option is B.