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

Options

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

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 follow these steps:

### Step 1: Change of Variable
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 as:

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

### Step 2: 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 3: Combine 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. \( I = \int_{a}^{b} \frac{f(a + b - x)}{f(a + b - x) + f(x)} \, dx \)

### Step 4: Add the Two Integrals
Adding both expressions for \( I \):

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

### Step 5: Simplify the Expression
The denominators are the same, so we can combine the fractions:

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

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

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

### Step 7: Solve for \( I \)
Thus, we have:

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

### Final Answer
The value of the integral is:

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

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