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If f(a + b + 1 − x) = f(x), for all x, where a and b are fixed positive real numbers, then 1/(a + b) ∫[a to b] x{f(x) + f(x + 1)} dx is equal to:

1. ∫[a+1 to b] f(x) dx
2. ∫[a−1 to b−1] f(x) dx
3. ∫[a−1 to b−1] f(x + 1) dx
4. ∫[a+1 to b+1] f(x + 1) dx

Correct answer: ∫[a−1 to b−1] f(x + 1) dx

Solution

The given functional equation indicates that the function f is symmetric about the point (a + b + 1)/2, which allows us to transform the integral by shifting the variable. This transformation leads to the conclusion that the integral of f(x + 1) over the interval [a−1, b−1] matches the original integral's structure, confirming that option C is the correct choice.

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