Re: [考古] 94成大
※ 引述《willmango (Tabo)》之銘言:
: 1
: 定積分∫x^1/2(1-x)^1/2dx
: 0
: 這題我跟同學討論了好久 用I.B.P.和令u 算出來都怪怪的
: 中間用三角代換 再換回來變成
: x=1
: ln(x^1/2)d│x=0
: 然後ln1 - ln0 => 失敗
: 懇請大大指教...Orz
1
∫ x^1/2(1-x)^1/2dx (set u=(1-x)^1/2 , dx = -2udu)
0
1 _____
= 2 ∫ √1-u^2 * u^2 du (set u=sinΘ , du=cosΘdΘ)
0
π/2 2 2
= 2 ∫ cos Θsin ΘdΘ
0
π/2
= (1/2) ∫ (1+cos2Θ)(1-cos2Θ) dΘ
0
π/2
= (1/2) ∫ (1 - (1/2)*(1+cos(4Θ) )dΘ
0
π/2 π/2 π/2
= (1/2) ∫ dΘ - (1/4) ∫ dΘ - (1/4) ∫ cos(4Θ)dΘ
0 0 0
= π/8 #
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