If `int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx=10`, then

Options

A. `b=22, a=2`
B. `b=15,a=-5`
C. `b=10, a=-10`
D. `b=10, a=-2`

Explanation

To solve the problem, we start with the given integral:

\[ I = \int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx = 10 \]

### Step 1: Use the property of definite integrals

We know from the properties of definite integrals that:

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

This means we can express the integral in a different way. Let's denote:

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

### Step 2: Rewrite the integral

Now, we can rewrite \(J\):

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

### Step 3: Add the two integrals

Now, we add \(I\) and \(J\):

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

### Step 4: Simplify the expression

Notice that:

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

Thus, we have:

\[ I + J = \int_{a}^{b} 1 \, dx = b - a \]

### Step 5: Relate \(I\) and \(J\)

Since \(I = J\) (because of the symmetry in the definitions of \(I\) and \(J\)), we can write:

\[ 2I = b - a \]

### Step 6: Solve for \(b - a\)

From the given information, we know that \(I = 10\):

\[ 2 \cdot 10 = b - a \]

Thus,

\[ b - a = 20 \]

### Conclusion

The difference between \(b\) and \(a\) is 20.

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