What is
`int_0^a (f(a -x))/(f(x) +f(a-x))dx`
equal to ?

Options

A. a
B. 2a
C. 0
D. `a/2`

Answer

To solve the integral

\[ I = \int_0^a \frac{f(a - x)}{f(x) + f(a - x)} \, dx, \]

we can use a property of definite integrals. This property states that

\[ \int_0^c f(x) \, dx = \int_0^c f(c - x) \, dx. \]

### Step 1: Set up the integral
Let’s denote the integral as \( I \):

\[ I = \int_0^a \frac{f(a - x)}{f(x) + f(a - x)} \, dx. \]

### Step 2: Change of variable
Now, we perform a change of variable. Let \( u = a - x \). Then, \( du = -dx \). When \( x = 0 \), \( u = a \) and when \( x = a \), \( u = 0 \). Thus, we can rewrite the integral as:

\[ I = \int_a^0 \frac{f(u)}{f(a - u) + f(u)} (-du) = \int_0^a \frac{f(u)}{f(a - u) + f(u)} \, du. \]

### Step 3: Rewrite the integral
Now, we can express this integral as:

\[ I = \int_0^a \frac{f(x)}{f(a - x) + f(x)} \, dx. \]

### Step 4: Add the two integrals
Now we have two expressions for \( I \):

1. \( I = \int_0^a \frac{f(a - x)}{f(x) + f(a - x)} \, dx \)
2. \( I = \int_0^a \frac{f(x)}{f(a - x) + f(x)} \, dx \)

Let’s add these two equations:

\[ 2I = \int_0^a \left( \frac{f(a - x)}{f(x) + f(a - x)} + \frac{f(x)}{f(a - x) + f(x)} \right) dx. \]

### Step 5: Simplify the expression
Notice that the two fractions inside the integral add up to 1:

\[ \frac{f(a - x)}{f(x) + f(a - x)} + \frac{f(x)}{f(a - x) + f(x)} = 1. \]

Thus, we have:

\[ 2I = \int_0^a 1 \, dx = \int_0^a dx = a. \]

### Step 6: Solve for \( I \)
Dividing both sides by 2 gives:

\[ I = \frac{a}{2}. \]

### Conclusion
Therefore, the value of the integral is:

\[ \int_0^a \frac{f(a - x)}{f(x) + f(a - x)} \, dx = \frac{a}{2}. \]