If `y=f(x)` satisfies has conditions of Rolle's theorem in `[2, 6],` then `int_(2)^(6)f'(x)dx` is equal to

Options

A. 2
B. 0
C. 4
D. 6

Answer

To solve the problem, we need to find the value of the integral \(\int_{2}^{6} f'(x) \, dx\) given that the function \(y = f(x)\) satisfies the conditions of Rolle's theorem on the interval \([2, 6]\).

### Step-by-Step Solution:

1. **Understanding Rolle's Theorem**:
Rolle's theorem states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \(f(a) = f(b)\), 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 \([2, 6]\). According to the conditions of Rolle's theorem:
- \(f(x)\) is continuous on \([2, 6]\).
- \(f(x)\) is differentiable on \((2, 6)\).
- \(f(2) = f(6)\).

3. **Setting Up the Integral**:
We need to evaluate the integral:
\[ \int_{2}^{6} f'(x) \, dx \]

4. **Using the Fundamental Theorem of Calculus**:
The Fundamental Theorem of Calculus states that:
\[ \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \]
Applying this to our integral, we have:
\[ \int_{2}^{6} f'(x) \, dx = f(6) - f(2) \]

5. **Substituting the Values**:
Since \(f(2) = f(6)\) (from the conditions of Rolle's theorem), we can substitute:
\[ f(6) - f(2) = k - k = 0 \]
where \(k\) is the common value of \(f(2)\) and \(f(6)\).

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

### Final Answer:
\[ \int_{2}^{6} f'(x) \, dx = 0 \]