`int_(a)^(b) (f(x))/(f(x) +f(a+b-x))dx`=

Options

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

Answer

To solve the integral

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

we can use a property of definite integrals. Let's go through the steps:

### Step 1: Define the Integral
Let

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

### Step 2: Change of Variable
Now, we will use the substitution \( x = a + b - t \). Then, when \( x = a \), \( t = b \) and when \( x = b \), \( t = a \). The differential \( dx = -dt \).

Thus, we can rewrite the integral:

\[ I = \int_{b}^{a} \frac{f(a+b-t)}{f(a+b-t) + f(t)} (-dt) = \int_{a}^{b} \frac{f(a+b-t)}{f(a+b-t) + f(t)} \, dt. \]

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

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

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

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

Adding these two equations gives:

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

### Step 5: Simplify the Expression
Notice that the denominators are the same, so we can combine the numerators:

\[ 2I = \int_{a}^{b} \frac{f(x) + f(a+b-x)}{f(x) + f(a+b-x)} \, dx = \int_{a}^{b} 1 \, dx. \]

### Step 6: Evaluate the Integral
The integral of 1 from \( a \) to \( b \) is simply:

\[ \int_{a}^{b} 1 \, dx = b - a. \]

Thus, we have:

\[ 2I = b - a. \]

### Step 7: Solve for \( I \)
Dividing both sides by 2 gives:

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

### Final Result
Therefore, the value of the integral is:

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

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