`f:[0,1)toR` be a non increasing function then for `alphaepsilon (0,1)`

Options

A. `alpha int_(0)^(1)f(x) dx le int_(0)^(alpha)f(x)dx`
B. `alpha int_(0)^(1)f(x)dx ge int_(0)^(alpha)f(x)dx`
C. `alpha^(2) int_(0)^(1)f(x)dxleint_(0)^(alpha) f(x)dx`
D. `sqrt(alpha) int_(0)^(1)f(x)dxgeint_(0)^(alpha)f(x)dx`

Explanation

To solve the problem, we need to analyze the given non-increasing function \( f: [0, 1) \to \mathbb{R} \) and the implications of the inequality involving the integral of \( f \).

### Step-by-step Solution:

1. **Understanding the Problem**: We are given that \( f \) is a non-increasing function defined on the interval \([0, 1)\). We need to analyze the inequalities involving the integral of \( f \) over different intervals.

2. **Change of Variables in the Integral**: We start with the integral \( \int_0^\alpha f(x) \, dx \). To analyze this, we can use a change of variables. Let \( x = \alpha t \), which implies \( dx = \alpha \, dt \). The limits of integration change accordingly: when \( x = 0 \), \( t = 0 \) and when \( x = \alpha \), \( t = 1 \).

\[ \int_0^\alpha f(x) \, dx = \int_0^1 f(\alpha t) \alpha \, dt \]

3. **Analyzing the Non-Increasing Property**: Since \( f \) is non-increasing, we have \( f(\alpha t) \geq f(t) \) for \( t \in [0, 1] \) because \( \alpha t < t \) when \( \alpha < 1 \).

4. **Integrating the Inequality**: We can now integrate both sides of the inequality \( f(\alpha t) \geq f(t) \):

\[ \int_0^1 f(\alpha t) \, dt \geq \int_0^1 f(t) \, dt \]

5. **Multiplying by \( \alpha \)**: Since \( \alpha > 0 \), we can multiply the entire inequality by \( \alpha \) without changing the direction of the inequality:

\[ \alpha \int_0^1 f(\alpha t) \, dt \geq \alpha \int_0^1 f(t) \, dt \]

6. **Substituting Back**: Recall that \( \int_0^\alpha f(x) \, dx = \alpha \int_0^1 f(\alpha t) \, dt \). Therefore, we can rewrite our inequality as:

\[ \int_0^\alpha f(x) \, dx \geq \alpha \int_0^1 f(x) \, dx \]

7. **Conclusion**: This means that the second option is correct:

\[ \int_0^\alpha f(x) \, dx \geq \alpha \int_0^1 f(x) \, dx \]

### Final Answer:
The correct option is:
\[ \int_0^\alpha f(x) \, dx \geq \alpha \int_0^1 f(x) \, dx \]