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), AA x in (0, 2)`, then F (2) equals :

Options

A. `e^(2)-1`
B. `e^(4)-1`
C. `e-1`
D. `e^(4)`

Answer

To solve the problem step by step, we start with the definitions and properties given in the question.

### Step 1: Define the function \( F(x) \)
We have:
\[ F(x) = \int_{0}^{x^2} f(\sqrt{t}) \, dt \]
where \( x \in [0, 2] \).

### Step 2: Differentiate \( F(x) \) using the Fundamental Theorem of Calculus
To find \( F'(x) \), we apply the Fundamental Theorem of Calculus and the chain rule:
\[ F'(x) = \frac{d}{dx} \left( \int_{0}^{x^2} f(\sqrt{t}) \, dt \right) = f(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = f(x) \cdot 2x \]
Thus, we have:
\[ F'(x) = 2x f(x) \]

### Step 3: Set up the equation given in the problem
We are given that \( F'(x) = f'(x) \) for \( x \in (0, 2) \). Therefore, we can equate:
\[ 2x f(x) = f'(x) \]

### Step 4: Rearranging the equation
Rearranging gives us a separable differential equation:
\[ \frac{f'(x)}{f(x)} = 2x \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = \int 2x \, dx \]
This leads to:
\[ \ln |f(x)| = x^2 + C \]
where \( C \) is the constant of integration.

### Step 6: Exponentiate to solve for \( f(x) \)
Exponentiating both sides gives:
\[ f(x) = e^{x^2 + C} = e^{C} e^{x^2} \]
Let \( e^{C} = k \), where \( k \) is a constant:
\[ f(x) = k e^{x^2} \]

### Step 7: Use the initial condition \( f(0) = 1 \)
We know from the problem statement that \( f(0) = 1 \):
\[ f(0) = k e^{0^2} = k \cdot 1 = k \]
Thus, \( k = 1 \) and we have:
\[ f(x) = e^{x^2} \]

### Step 8: Find \( F(2) \)
Now we need to compute \( F(2) \):
\[ F(2) = \int_{0}^{2^2} f(\sqrt{t}) \, dt = \int_{0}^{4} f(\sqrt{t}) \, dt \]
Substituting \( f(\sqrt{t}) = e^{(\sqrt{t})^2} = e^{t} \):
\[ F(2) = \int_{0}^{4} e^{t} \, dt \]

### Step 9: Evaluate the integral
The integral of \( e^{t} \) is:
\[ \int e^{t} \, dt = e^{t} + C \]
Thus:
\[ F(2) = \left[ e^{t} \right]_{0}^{4} = e^{4} - e^{0} = e^{4} - 1 \]

### Conclusion
Therefore, the value of \( F(2) \) is:
\[ F(2) = e^{4} - 1 \]