→ k6416337 :根據黎曼可積的定義下，有這個定理結果嗎? 11/29 09:03
→ egg12388 :upper & lower sum? 12/03 17:29
取partition
x0=1
1 1
xn=(── + ──)/2 n>=1
n n+1
x0'=0
then x0>x1>x2>....>xn>x0' (中間砍n刀的情況)
Now consider finite Riemann sum
n
Σ f(x)(x - x ) + f(0)(x - x ) (最後一項只是為了補足整個定義域)
k=1 k k k-1 0' n
is equal to 0 , because for k from 1 to n , f(xk) =0 , and f(0)=0
so when n→infinity
n
lim Σ f(x)(x - x ) + f(0)(x - x ) = 0 (最後一項只是為了補足整個定義域)
n→inf k=1 k k k-1 0' n
so the Riemann integral of f(x) exists , equal to 0
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◆ From: 114.25.179.188
※ 編輯: znmkhxrw 來自: 114.25.179.188 (11/29 03:22)