Re: [商管] 求重積分機率

看板Grad-ProbAsk作者 (廣伸小隊正夯!!)時間16年前 (2010/03/24 02:24), 編輯推噓7(709)
留言16則, 3人參與, 最新討論串1/1
※ 引述《hide5209 (Drugs can't kill teens )》之銘言: :   1 1 ^0.5 :  ∫ ∫ (X^2+Y^2) 4*X*YdXdY :   0 0  : 求機率 : 可以板上的同學幫忙解一下嗎?? 我硬積= = 1 1 S 4y S ((x^2+y^2)^0.5)*x dx dy (let u=x^2+y^2, then du=2x*dx) 0 0 1 (1+y^2) => S 2y S u^0.5 du dy 0 (y^2) 1 1+y^2 =>S 2y [-2u^(-0.5)] dy 0 y^2 1 -4y =>S (------------ +4) dy (let k=1+y^2, then dk=2y*dy) 0 (1+y^2)^0.5 2 1 =>-2*S (k^-0.5)dk +[4y] 1 0 2 =>-2*[2*(k)^0.5] +4 = 2.343# (大概算一下,不知道對不對...) 1 -- HEY~~~HEY~~~CHAAAAARLIE~!   。。。。。 ⊙ . ▼▼▼▼ \▲▲▲▲ φ . \ . δ ./ㄨ \\/ˊ▄▄ \|/ㄑ ( ︶ ˋ\///\/. by Armour@joke -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.114.231.103 ※ 編輯: loveliver 來自: 140.114.231.103 (03/24 02:27)

03/24 02:29, , 1F
答案是0.9752,但是我不管怎麼算都大於1阿
03/24 02:29, 1F

03/24 02:38, , 2F
我算出來是 (2√2 - 1)8/15
03/24 02:38, 2F

03/24 02:45, , 3F
d大可以寫詳解給我嗎
03/24 02:45, 3F

03/24 02:48, , 4F
= ∫2y ∫ √(x^2+y^2) d(x^2+y^2) dy
03/24 02:48, 4F

03/24 02:49, , 5F
= ∫(4/3)y[(1 + y^2)^(3/2) - y^3] dy
03/24 02:49, 5F

03/24 02:49, , 6F
那裡為什麼是2Y不是4Y先提出來嗎
03/24 02:49, 6F

03/24 02:52, , 7F
因為 2xdx = d(x^2) , 但x和y是 independent
03/24 02:52, 7F

03/24 02:52, , 8F
所以 d(x^2) = d(x^2+y^2)
03/24 02:52, 8F

03/24 02:56, , 9F
2xd2=dx^2 為什麼不是2dx^2
03/24 02:56, 9F

03/24 02:57, , 10F
d(x^2)/dx = 2x → d(x^2) = (2x)dx
03/24 02:57, 10F

03/24 03:00, , 11F
太感謝了..我瞭了
03/24 03:00, 11F
※ 編輯: loveliver 來自: 140.114.231.103 (03/24 03:10)

03/24 03:08, , 12F
我算到2/3S{2/5(1+y^2)^5/2-y^3用Y^2+1來積但是不知道那
03/24 03:08, 12F

03/24 03:09, , 13F
y^3改怎麼辦
03/24 03:09, 13F

03/24 03:10, , 14F
= =好像還是不對,參考就好...
03/24 03:10, 14F

03/24 03:13, , 15F
lim{a+b} = lim{a} + lim{b} , 把那兩團拆開來積分
03/24 03:13, 15F

03/24 03:30, , 16F
總於算出來呵阿~~D大~~~L大感謝你們
03/24 03:30, 16F
文章代碼(AID): #1BgGTwBn (Grad-ProbAsk)