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Question
Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx` if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.
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Solution
Let `I = int_0^a f(x) g(x) dx`
`= int_0^a f(a - x) [4 - g(a - x)] dx`
`= 4 int_0^a f(a - x) dx - int_0^a f(a - x) g (a - x) dx`
Let a - x = t
⇒ - dx = dt
When x = 0, t = a
and x = a, t = 0
`I = -4 int_a^0 f (t) dt + int_a^0 f (t) g (t) dt`
`= 4 int_0^a f (t) dt - int_0^a f (t) g (t) dt`
`= 4 int_0^a f (x) dx - int_0^a f (x)g (x) dx `
`= 4 int_0^a f (x) dx - I`
⇒ `2I = 4 int_0^a f (x) dx`
Hence, `I = 2 int_0^a f (x) dx`
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