若函數ƒ(x)在[0,1]上連續可微,證明：
lim(n→∞) n∫(0~1) xⁿƒ(x) dx = ƒ(1)
定積分證明
2012-05-17 1:39 am
回答 (2)
2012-05-17 2:29 am
✔ 最佳答案
∫(0~1) nx^nƒ(x) dx =nf(1)/(n+1) -∫(0~1) n/(n+1)x^(n+1)ƒ'(x) dx (integration by parts)
=nf(1)/(n+1) - f'(k) ∫(0~1) n/(n+1) x^(n+1) dx (M.V.T. for definite integration for some k)
=nf(1)/(n+1) - f'(k) n/[(n+1)(n+2)]
so
lim(n→∞) n∫(0~1) xⁿƒ(x) dx
=lim(n→∞) nf(1)/(n+1) - lim(n→∞) f'(k) n/[(n+1)(n+2)]
=f(1)
2012-05-20 9:29 am
請注意 k depends on n, but it does not matter, since f' is bounded on [0, 1].
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120516000051KK00479