Let f(x) be a differentiable function in the interval (0,2), then the value of ∫20f(x)dxf(c) where c∈(0,2)2f(c) where c∈(0,2)f′(c) where c∈(0,2)None of these

To analyze the problem and find the answer, we will use the Mean Value Theorem for definite integrals. Here's a structured breakdown of the solution.

1. Break Down the Problem

We need to evaluate the integral $\int_0^2 f(x) \, dx$ and relate it to the function values and derivatives at some point $c \in (0, 2)$.

2. Relevant Concepts

The Mean Value Theorem for Integrals states that if $f(x)$ is continuous on $[a, b]$, then there exists a point $c$ in $(a, b)$ such that:

$\int_a^b f(x) \, dx = f(c) \cdot (b - a)$

In our case, $a = 0$ and $b = 2$. Therefore:

$\int_0^2 f(x) \, dx = f(c) \cdot (2 - 0) = 2f(c)$

3. Analysis and Detail

Since $f(x)$ is a differentiable function on $(0, 2)$, we can apply the theorem directly. This gives us:

1. $\int_0^2 f(x) \, dx = 2f(c)$

Now, we also need to consider $2f(c)f'(c)$. According to the Mean Value Theorem and differentiability, $f'(c)$ represents the average rate of change of $f$ at $c$.

4. Verify and Summarize

From the result derived, we can confirm that:

$\int_0^2 f(x) \, dx = 2f(c)$

And under the context of provided options (including 2f(c), f′(c) and so forth), only $2f(c)$ falls directly from the conclusion of the integral's evaluation.

Final Answer

The value of $\int_0^2 f(x) \, dx$ is $2f(c)$ for some $c \in (0, 2)$. Thus, the correct option is:

2f(c)