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.
高微(integrable and mean value)
2012-11-16 3:41 am
回答 (2)
2012-11-17 4:38 pm
✔ 最佳答案
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。
2012-11-16 6:18 am
取 x_1 = a, 用中間值定理於 f. (設 g 的定積分大於 0)
收錄日期: 2021-05-04 01:50:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121115000015KK05494