If f(x)=f(a+b-x) for all `x in[a,b]` and `int_(a)^(b) xf(x) dx=k int_(a)^(b) f(x) dx`, then the value of k, is

Options

A. `(a+b)/(2)`
B. `(a-b)/(2)`
C. `(a^(2)+b^(2))/(2)`
D. `(a^(2)-b^(2))/(2)`

Answer

To solve the problem, we start with the given function and the integral equation.

### Step 1: Understand the given condition
We have the function \( f(x) = f(a + b - x) \) for all \( x \) in the interval \([a, b]\). This property indicates that the function is symmetric about the midpoint \( \frac{a + b}{2} \).

**Hint:** Recognize that this symmetry can be leveraged to transform the integral.

### Step 2: Set up the integral
Let \( I = \int_a^b x f(x) \, dx \). We will apply the property of definite integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
to our integral.

**Hint:** Identify how to express \( I \) using the symmetry property of \( f(x) \).

### Step 3: Apply the symmetry property
Using the property, we can rewrite the integral:
\[ I = \int_a^b x f(x) \, dx = \int_a^b (a + b - x) f(a + b - x) \, dx \]
By substituting \( u = a + b - x \), we find that \( du = -dx \), which changes the limits of integration accordingly.

**Hint:** Remember to change the limits of integration when substituting variables.

### Step 4: Change the variable in the integral
When we change the variable, we have:
\[ I = \int_b^a (a + b - u) f(u) (-du) = \int_a^b (a + b - u) f(u) \, du \]
This can be rewritten as:
\[ I = \int_a^b (a + b) f(u) \, du - \int_a^b u f(u) \, du \]
This simplifies to:
\[ I = (a + b) \int_a^b f(x) \, dx - I \]

**Hint:** Notice how \( I \) appears on both sides of the equation.

### Step 5: Solve for \( I \)
Now, we can combine the terms:
\[ 2I = (a + b) \int_a^b f(x) \, dx \]
Thus, we have:
\[ I = \frac{(a + b)}{2} \int_a^b f(x) \, dx \]

**Hint:** This expression matches the form given in the problem statement.

### Step 6: Relate to the given equation
From the problem, we know:
\[ I = k \int_a^b f(x) \, dx \]
By comparing both expressions for \( I \), we find:
\[ k = \frac{(a + b)}{2} \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{\frac{a + b}{2}} \]