Re: [微積] 連續函數的積分與極限一題
※ 引述《cxcxvv (delta)》之銘言：
: Suppose f is continuous on [0,1]
: Define
: 1
: Xn = (n+1)∫x^n f(x) dx
: 0
: Find lim Xn
: n->infinity
: 原本以為用積分中值定理就做出來了
: 可是那個c belongs to (0,1)會depend on n
: 所以其實不能那樣做
: 那這一題要怎麼解呢?
Another way:
0, x in [0,1)
Let A_n(x) = x^(n+1). Then lim A_n(x) = A(x) = {
n→∞ 1, x = 1
1 1
And (n+1)∫ x^n f(x) dx = ∫f(x) d(x^(n+1))
0 0
1
= ∫f(x) d(A_n(x))
0
1 1
Hence, lim ∫f(x) d(A_n(x)) = ∫f(x) d(A(x)) = f(1)
n→∞ 0 0
Q.E.D.
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※ 編輯: CFE220 來自: 218.170.114.141 (01/19 18:55)
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