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Let f and g be continuous on the interval [0,a], with f(a−x)=f(x) and g(x)+g(a−x)=2 for every x in [0,a]. Then the value of ∫₀^a f(x)g(x) dx is:

1. ∫₀^a f(x) dx
2. ∫₀^a g(x) dx
3. 0
4. None of these

Correct answer: ∫₀^a f(x) dx

Solution

Let I=int_0^a f(x)g(x)dx. Substituting x->a-x and using f(a-x)=f(x), g(a-x)=2-g(x): I=int_0^a f(x)(2-g(x))dx = 2*int_0^a f dx - I. So 2I=2*int_0^a f dx, giving I=int_0^a f(x)dx.

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