maths definite integral
2011-04-28 2:00 am
1. Let f(x) be a function. If g(x) is a periodic function with period 1/2 and g(x)= ∫(t=0 to x) (f(t)-2k) dt, where k is a constant, show that k= ∫(t= 0 to 1/2) f(t) dt
回答 (1)
2011-04-28 5:14 am
✔ 最佳答案
g(x)= ∫(t=0 to x) (f(t)-2k) dtg(x + 1/2) = g(x)
∫(t=0 to x + 1/2) (f(t)-2k) dt = ∫(t=0 to x) (f(t)-2k) dt
∫(t=x to x + 1/2) (f(t)-2k) dt = 0
So ∫(t=x to x + 1/2) f(t) dt = ∫(t=x to x + 1/2) 2k dt
2k(x + 1/2 - x) = ∫(t=x to x + 1/2) f(t) dt
k = ∫(t=x to x + 1/2) f(t) dt
Since g(x) is periodic and we can take x = 0 such that k = ∫(t=0 to 1/2) f(t) dt
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