Let `f(x) and g(x)` be any two continuous function in the interval `[0, b]` and 'a' be any point between 0 and b. Which satisfy the following conditions : `f(x)=f(a-x), g(x)+g(a-x)=3, f(a+b-x)=f(x)`. Also `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx, int_(a)^(b)f(x)dx=int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`
If `f(a+b-x)=f(x)`, then `int_(a)^(b)xf(x)dx` is

Options

A. `(a+b)/(2)int_(a)^(b)f(x)dx`
B. `(a-b)/(2)int_(a)^(b)f(x)dx`
C. 0
D. none of these

Answer

To solve the problem, we need to analyze the given conditions and apply them step by step.

### Step 1: Understand the properties of the functions
We have two continuous functions \( f(x) \) and \( g(x) \) defined on the interval \([0, b]\) with certain symmetry properties:
1. \( f(x) = f(a - x) \)
2. \( g(x) + g(a - x) = 3 \)
3. \( f(a + b - x) = f(x) \)

### Step 2: Analyze the integrals
We are given the following integral conditions:
1. \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \)
2. \( \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \)

From the first condition, since \( f(x) = f(a - x) \), we can conclude that the integral from \( 0 \) to \( a \) is symmetric.

### Step 3: Evaluate the integral \( \int_a^b x f(x) \, dx \)
Using the property \( f(a + b - x) = f(x) \):
\[ \int_a^b x f(x) \, dx = \int_a^b (a + b - x) f(a + b - x) \, dx \]
Substituting \( u = a + b - x \) (which implies \( du = -dx \)):
\[ \int_a^b x f(x) \, dx = \int_b^a (a + b - u) f(u) (-du) = \int_a^b (a + b - u) f(u) \, du \]
This gives us:
\[ \int_a^b x f(x) \, dx = \int_a^b (a + b) f(u) \, du - \int_a^b u f(u) \, du \]

### Step 4: Combine the integrals
Let \( I = \int_a^b x f(x) \, dx \). Then we have:
\[ I = (a + b) \int_a^b f(x) \, dx - I \]
This leads to:
\[ 2I = (a + b) \int_a^b f(x) \, dx \]
Thus:
\[ I = \frac{(a + b)}{2} \int_a^b f(x) \, dx \]

### Conclusion
The value of \( \int_a^b x f(x) \, dx \) is given by:
\[ \int_a^b x f(x) \, dx = \frac{(a + b)}{2} \int_a^b f(x) \, dx \]