If `f(x)` satisfies the condition of Rolle's theorem in `[1,2]`, then `int_1^2 f'(x) dx` is equal to (a) 1 (b) 3 (c) 0 (d) none of these

Options

A. `1`
B. `3`
C. `0`
D. none of these

Explanation

To solve the problem, we need to analyze the integral of the derivative of the function \( f(x) \) over the interval \([1, 2]\) given that \( f(x) \) satisfies the conditions of Rolle's Theorem.

### Step-by-Step Solution:

1. **Understanding Rolle's Theorem**:
Rolle's Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

2. **Applying the Conditions**:
In our case, we have the interval \([1, 2]\). According to the problem:
- \( f(x) \) is continuous on \([1, 2]\),
- \( f(x) \) is differentiable on \((1, 2)\),
- \( f(1) = f(2) \).

3. **Setting Up the Integral**:
We need to evaluate the integral:
\[ \int_1^2 f'(x) \, dx \]

4. **Using the Fundamental Theorem of Calculus**:
By the Fundamental Theorem of Calculus, we can express the integral of the derivative as:
\[ \int_1^2 f'(x) \, dx = f(2) - f(1) \]

5. **Substituting the Values**:
Since we know from the conditions of Rolle's Theorem that \( f(1) = f(2) \), we can substitute this into our equation:
\[ f(2) - f(1) = 0 \]

6. **Conclusion**:
Therefore, we find that:
\[ \int_1^2 f'(x) \, dx = 0 \]

### Final Answer:
Thus, the value of the integral \( \int_1^2 f'(x) \, dx \) is \( 0 \). The correct option is (c) 0.