The function `f` and `g` are positive and continuous. If `f` is increasing and `g` is decreasing, then `int_0^1f(x)[g(x)-g(1-x)]dx` is always non-positive is always non-negative can take positive and negative values none of these

Options

A. is always non-positive
B. is always non-negative
C. can take positive and negative values
D. none of these

Explanation

To solve the problem, we need to analyze the integral

\[ I = \int_0^1 f(x) [g(x) - g(1-x)] \, dx \]

where \( f \) is an increasing function and \( g \) is a decreasing function. We will show that this integral is always non-positive.

### Step 1: Rewrite the Integral
We can use the property of definite integrals:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, we can rewrite the integral as:

\[ I = \int_0^1 f(1-x) [g(1-x) - g(x)] \, dx \]

### Step 2: Change of Variables
Now, let's perform a change of variable \( u = 1 - x \). When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 0 \). Thus, the integral becomes:

\[ I = \int_1^0 f(u) [g(u) - g(1-u)] (-du) = \int_0^1 f(u) [g(u) - g(1-u)] \, du \]

### Step 3: Combine the Integrals
Now, we have two expressions for \( I \):

1. \( I = \int_0^1 f(x) [g(x) - g(1-x)] \, dx \)
2. \( I = \int_0^1 f(1-x) [g(1-x) - g(x)] \, dx \)

Adding these two expressions for \( I \):

\[ 2I = \int_0^1 \left( f(x) - f(1-x) \right) [g(x) - g(1-x)] \, dx \]

### Step 4: Analyze the Terms
- Since \( f \) is increasing, \( f(1-x) \geq f(x) \) when \( x \in [0, 0.5] \) and \( f(1-x) \leq f(x) \) when \( x \in [0.5, 1] \).
- Since \( g \) is decreasing, \( g(x) \geq g(1-x) \) when \( x \in [0, 0.5] \) and \( g(x) \leq g(1-x) \) when \( x \in [0.5, 1] \).

### Step 5: Determine the Sign of \( I \)
1. For \( x \in [0, 0.5] \):
- \( f(x) - f(1-x) \leq 0 \)
- \( g(x) - g(1-x) \geq 0 \)
- Thus, \( (f(x) - f(1-x))(g(x) - g(1-x)) \leq 0 \)

2. For \( x \in [0.5, 1] \):
- \( f(x) - f(1-x) \geq 0 \)
- \( g(x) - g(1-x) \leq 0 \)
- Thus, \( (f(x) - f(1-x))(g(x) - g(1-x)) \leq 0 \)

### Conclusion
In both intervals, the product \( (f(x) - f(1-x))(g(x) - g(1-x)) \) is non-positive. Therefore, we conclude that:

\[ 2I \leq 0 \implies I \leq 0 \]

Thus, the integral \( I \) is always non-positive.

### Final Answer
The correct option is: **always non-positive**.

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