推 mikshone:厲害...謝謝 163.22.18.105 05/15
※ 引述《mikshone (HIHI)》之銘言：
: 1 t
: 1. f(x)=x + S f(t)e dt 求f(x)
: 0
1
∫:=∫
0
since f'(x) = 1,f(x) = x + c;where ∫ f(t)e^{t}dt = c
c =∫f(t)e^{t}dt = ∫(t+c)e^{t}dt = ce-c+1 ==> c = 1/(2-e)
thus f(x) = x + 1/(2-e)
1
: x -1 4
: 2. S xf(t)dt=S f(t)dt + x + 3 求f(x)
: 1 x
: 5 34 -1
: 3. f(x)= x + x 求S f(x) dx
: 2
34 2
∫:=∫ ∫:=∫
2 1
let y = f(x),then x = f^{-1}y
thus ∫f^{-1}ydy = ∫xd(x^5 + x)
= ∫x(5x^4+1)dx = 54
: 1 2
: Ans: 1.x + ----- 2. 3x -2x +1 3.54
: 2-e
: 幫忙一下 謝謝啦.....
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◆ From: 61.58.162.122
※ 編輯: okhunter 來自: 61.58.162.122 (05/15 00:14)