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# 高微(integrable and mean value)

Show that if f and g are nonnegative real functions on [a, b], with f

continuous on [a, b] and g integrable on [a, b], then there exist

x_0, x_1[∈[a, b] such that

∫(a to b) f(x)g(x)dx = f(x_0)∫(x_1 to b) g(x)dx.

### 2 Answers

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- 感冒唔駛食藥，飲盒仔茶就得啦。Lv 57 years agoFavorite Answer
由於f和g非負，固对所有c，a≤c≤b，

∫(a to c) fg≥0

設x_1=inf{c: ∫(a to c) fg >0 }，即有

∫(a to b) fg = ∫(x_1 to b) fg

且

min{f(t): x_1≤t≤b }∫(x_1 to b) fg

≤ ∫(x_1 to b) fg

≤ max{f(t): x_1≤t≤b }∫(x_1 to b) fg

由連續函数的中值定理，存在x_0使得

∫(x_1 to b) fg=f(x_0)∫(x_1 to b) g, x_1≤x_0≤b

得証。

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