Integral Defined Function and Derivative Relation

To find the value of $F(2)$, we begin by simplifying the expression for $F(x)$.

The function $F(x)$ is defined for $x \in [0, 2]$ as: $F(x) = \int_{0}^{x^2} f(\sqrt{t}) \, dt$

We can evaluate the derivative $F'(x)$ using Leibniz's rule for differentiation under the integral sign: $F'(x) = \frac{d}{dx}\left(\int_{0}^{x^2} f(\sqrt{t}) \, dt\right) = f\left(\sqrt{x^2}\right) \cdot \frac{d}{dx}(x^2) - 0$ Since $x \in (0, 2)$, we have $\sqrt{x^2} = x$. Therefore: $F'(x) = 2x f(x)$

Alternatively, using the substitution $t = u^2$ (with $dt = 2u \, du$): $F(x) = \int_{0}^{x} 2u f(u) \, du$ By the Fundamental Theorem of Calculus: $F'(x) = 2x f(x)$

We are given that: $F'(x) = f'(x) \quad \text{for all } x \in (0, 2)$

Equating the two expressions for $F'(x)$: $f'(x) = 2x f(x)$

Dividing both sides by $f(x)$ (since $f(0) = 1 \neq 0$ and $f$ is continuous, $f(x) > 0$ on $[0, 2]$): $\frac{f'(x)}{f(x)} = 2x$

Integrating both sides with respect to $x$: $\int \frac{f'(x)}{f(x)} \, dx = \int 2x \, dx$ $\ln |f(x)| = x^2 + C$ $f(x) = A e^{x^2}$

Using the given initial condition $f(0) = 1$: $f(0) = A e^0 = 1 \implies A = 1$ Thus, $f(x) = e^{x^2}$

Now, since $F'(x) = f'(x)$ on $(0, 2)$, integrating both sides from $0$ to $x$ yields: $\int_{0}^{x} F'(t) \, dt = \int_{0}^{x} f'(t) \, dt$ $F(x) - F(0) = f(x) - f(0)$

From the definition of $F(x)$, we have: $F(0) = \int_{0}^{0} f(\sqrt{t}) \, dt = 0$

Thus: $F(x) = f(x) - f(0) = e^{x^2} - 1$

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

Therefore, the correct option is (B).