If f(x) satisfies the requirements of Rolle's Theorem in [1,2] and f(x) is continuous in [1,2] then ` int_(1)^(2) f'(x)` dx is equal to

Options

A. 0
B. 1
C. 3
D. `-1`

Answer

To solve the problem, we need to evaluate the integral of the derivative of a 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 the Integral**:
We need to evaluate the integral:
\[ I = \int_{1}^{2} f'(x) \, dx \]

2. **Applying the Fundamental Theorem of Calculus**:
According to the Fundamental Theorem of Calculus, if \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then:
\[ \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \]
Here, \( a = 1 \) and \( b = 2 \).

3. **Evaluating the Integral**:
Applying the theorem, we have:
\[ I = f(2) - f(1) \]

4. **Using Rolle's Theorem**:
Since \( f(x) \) satisfies the conditions of Rolle's Theorem on the interval \([1, 2]\), it implies that:
\[ f(1) = f(2) \]

5. **Substituting into the Integral**:
Therefore, substituting \( f(2) = f(1) \) into our expression for \( I \):
\[ I = f(2) - f(1) = f(1) - f(1) = 0 \]

6. **Conclusion**:
Thus, the value of the integral is:
\[ \int_{1}^{2} f'(x) \, dx = 0 \]

### Final Answer:
The value of the integral \( \int_{1}^{2} f'(x) \, dx \) is \( 0 \).

---