→ young11539 :! KEY是令Y = PI -X 06/05 18:17
→ young11539 :THX 06/05 18:18
※ 引述《young11539 (〝☆小小霈★”)》之銘言：
: if f(x) conti. on [0,1]
: then show that
: ∫xf(sinx)dx = (π/2)∫f(sinx)dx
: 積分範圍 [0,π]
: 感激不盡
設 x=π-y, dx=-dy
π
∫ xf(sinx)dx
0
0
= ∫(π-y)f(sin(π-y))(-dy)
π
π
= ∫(π-y)f(siny)dy
0
π π
= π∫ f(siny)dy - ∫yf(siny)dy
0 0
∵變數不影響積分值
π π π
∴∫ xf(sinx)dx = π∫ f(sinx)dx - ∫xf(sinx)dx
0 0 0
π π
=>∫ xf(sinx)dx = (π/2)∫ f(sinx)dx #
0 0
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※ 編輯: aegius1r 來自: 111.250.57.224 (06/05 16:51)