Let f(x) be a differentiable function in the interval (0, 2) then the value of `int_(0)^(2)f(x)dx`

Options

A. f(c) where `c in (0,2)`
B. `2f(c),` where `c in (0, 2)`
C. `f'(c)` where `c in (0, 2)`
D. None of these

Answer

To solve the problem, we need to evaluate the definite integral of a differentiable function \( f(x) \) over the interval \((0, 2)\). We will use the Mean Value Theorem for integrals to find the value of the integral.

### Step-by-step Solution:

1. **Understanding the Integral**:
We need to evaluate the integral \( \int_0^2 f(x) \, dx \).

2. **Applying the Mean Value Theorem for Integrals**:
The Mean Value Theorem for integrals states that if \( f(x) \) is continuous on the interval \([a, b]\), then there exists at least one \( c \in (a, b) \) such that:
\[ \int_a^b f(x) \, dx = f(c) \cdot (b - a) \]
In our case, \( a = 0 \) and \( b = 2 \).

3. **Substituting Values**:
Applying the theorem, we have:
\[ \int_0^2 f(x) \, dx = f(c) \cdot (2 - 0) = 2f(c) \]
where \( c \) is some value in the interval \( (0, 2) \).

4. **Conclusion**:
Therefore, the value of the integral \( \int_0^2 f(x) \, dx \) can be expressed as:
\[ \int_0^2 f(x) \, dx = 2f(c) \]
where \( c \) is in the interval \( (0, 2) \).

### Final Answer:
The value of \( \int_0^2 f(x) \, dx \) is \( 2f(c) \) for some \( c \in (0, 2) \).

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