We are given that \( f \) is differentiable on the interval \([0, 1]\), and we need to analyze the given statements.
Step 1: Consider the integral and the point \( c \)
We are interested in the integral:
\[
\int_0^c f(x) \, dx
\]
and we are asked whether there exists a point \( c \in (0, 1) \) such that the following relationship holds:
\[
\int_0^c f(x) \, dx = (1 - c) f(c)
\]
This question is related to the application of the **Mean Value Theorem for Integrals**. The Mean Value Theorem for Integrals states that if \( f \) is continuous on \([a, b]\), then there exists a point \( c \in [a, b] \) such that:
\[
\int_a^b f(x) \, dx = f(c)(b - a)
\]
Here, we can apply this theorem to the interval \([0, c]\), which gives us:
\[
\int_0^c f(x) \, dx = f(c) \cdot c
\]
So, the equation we are checking becomes:
\[
f(c) \cdot c = (1 - c) \cdot f(c)
\]
Simplifying this, we get:
\[
c = (1 - c)
\]
which implies:
\[
c = \frac{1}{2}
\]
Thus, there exists a point \( c \in (0, 1) \) where the integral satisfies the given relationship. Therefore, the first statement is correct.
Step 2: Analyze the other options
- The second option says that there does not exist any point \( d \in (0, 1) \) for which the relationship holds. This is incorrect, as we showed that \( c = \frac{1}{2} \) satisfies the equation.
- The third option states that \( \int_0^c f(x) \, dx \) does not exist for any \( c \in (0, 1) \). This is not true, as the integral is well-defined for any differentiable function \( f \) on \([0, 1]\).
- The fourth option suggests that \( \int_0^c f(x) \, dx \) is independent of \( c \), which is clearly incorrect since the value of the integral depends on the upper limit of integration, \( c \).
Conclusion
The correct answer is:
\[
\boxed{\text{There exists } c \in (0, 1) \text{ such that } \int_0^c f(x) \, dx = (1 - c) f(c)}
\]