F.5 M2 integration~$%&~

2011-05-08 10:08 pm

It is given that f(x) is a continuous function, where its domain is all real numbers. Define p(x) = f(x) - f(-x). Prove that ∫ [Lower limit: -a , upper limit: a] p(x)dx = 0 , where a > 0.

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用戶25975

2011-05-08 10:18 pm

✔ 最佳答案

Consider p(-x) = f(-x) - f(x) = -p(x)

We have p(x) is an odd function and therefore:

∫ (x = -a → a) p(x) dx = ∫ (x = -a → 0) p(x) dx + ∫ (x = 0 → a) p(x) dx

For ∫ (x = -a → 0) p(x) dx, we sub u = -x and then du = -dx.

When x = 0, u = 0 and when x = -a, u = a. Hence:

∫ (x = -a → 0) p(x) dx = - ∫ (u = a → 0) p(-u) du

= ∫ (u = 0 → a) p(-u) du

= - ∫ (x = 0 → a) p(x) dx

Finally:

∫ (x = -a → a) p(x) dx = ∫ (x = -a → 0) p(x) dx + ∫ (x = 0 → a) p(x) dx

= - ∫ (x = 0 → a) p(x) dx + ∫ (x = 0 → a) p(x) dx

= 0

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