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[積分] 幾題問題

[ trans_math ]15 留言, 推噓總分: +2

作者: blak - 發表於 2012/06/16 21:28(12年前)

1^F推suhorng:第二題把cos(t)的泰勒展開中t用x^2帶入61.217.32.242 06/16 21:38

2^F→suhorng:3.令t=1/x, dx/x^2 = -dt61.217.32.242 06/16 21:39

4^F→suhorng:yes~61.217.32.242 06/16 22:40

6^F→suhorng:沒有算...XD61.217.32.242 06/16 22:42

[考古] 台大100年為B計算第二題

[ trans_math ]8 留言, 推噓總分: +2

作者: make990 - 發表於 2012/06/14 20:44(12年前)

1^F推suhorng:1-cos(x) = 2sin^2(x/2), let u=x/261.217.32.242 06/14 20:47

2^F→suhorng:所謂簡化應該是說變成這樣後可以代公式吧?61.217.32.242 06/14 20:47

4^F→suhorng:第一行代進去那樣變數變換則1024=64*8*261.217.32.242 06/14 21:27

5^F→suhorng:算一下阿61.217.32.242 06/14 21:27

7^F推suhorng:令 u=x/2, 則 dx = 2 du <-- 這裡來的61.217.32.242 06/14 22:19

Re: [積分]台大100年微B第五題

[ trans_math ]16 留言, 推噓總分: +6

作者: PaulErdos - 發表於 2012/06/08 21:51(12年前)

8^F→suhorng:No, f是被積函數, 積分範圍選單位球59.115.148.117 06/09 09:08

Re: [積分] 重積分

[ trans_math ]16 留言, 推噓總分: +1

作者: BaBi - 發表於 2012/06/07 00:17(12年前)

3^F→suhorng:y是變數x是常數61.217.35.109 06/07 08:23

9^F→suhorng:是吧61.217.35.109 06/07 09:05

[積分] 重積分

[ trans_math ]1 留言, 推噓總分: 0

作者: ApplebyArrow - 發表於 2012/06/07 00:05(12年前)

1^F→suhorng:...嗯? xe^(x^2/x) = xe^x 是這樣嗎...61.217.35.109 06/07 00:16

[積分] 積分

[ trans_math ]20 留言, 推噓總分: +3

作者: blak - 發表於 2012/06/05 17:16(12年前)

1^F推suhorng:1.內=6(sec^2(2x)-1)*tan sec,令t=sec(2x)61.217.35.142 06/05 17:18

2^F→suhorng:第二題的積分範圍會畫嗎? z積完用極座標.61.217.35.142 06/05 17:19

3^F→suhorng:(直接用原柱座標是一樣意思)61.217.35.142 06/05 17:20

5^F→suhorng:令 t = sec(2x) 代換61.217.35.142 06/05 17:42

7^F→suhorng:我代換完是變成∫3(t^2-1)dt @@61.217.35.142 06/05 17:43

9^F→suhorng:(sec 2x)' = sec 2x tan 2x61.217.35.142 06/05 17:45

10^F→suhorng: * 261.217.35.142 06/05 17:48

13^F→suhorng:3(sec^2(2x) - 1)(sec 2x)' = 6tan^3 sec61.217.35.142 06/05 17:56

14^F→suhorng:(sec 2x)' = sec 2x tan 2x * 261.217.35.142 06/05 17:56

15^F→suhorng:對61.217.35.142 06/05 17:56

19^F推suhorng:4 因為6會變361.217.35.109 06/06 08:19

[微分] 偏微

[ trans_math ]6 留言, 推噓總分: 0

作者: blak - 發表於 2012/06/05 15:12(12年前)

1^F→suhorng:yes61.217.35.142 06/05 17:33

[羅必達] 請問這三題為何能用羅必達

[ trans_math ]18 留言, 推噓總分: +2

作者: rad80126 - 發表於 2012/06/04 21:07(12年前)

3^F→suhorng:Yes, you are right. Now note that61.217.35.142 06/05 15:53

4^F→suhorng:lim_{u→0} sin(t)/t = 161.217.35.142 06/05 15:53

5^F→suhorng:So for N large enough, we may replace61.217.35.142 06/05 15:53

6^F→suhorng:∫[N,x]sin(1/(1+t))dt by∫[N,x]dt/(1+t)61.217.35.142 06/05 15:54

7^F→suhorng:which diverges to ∞ as x→∞.61.217.35.142 06/05 15:54

8^F推suhorng:For 3, cos(t)/t^2≧cos(1/2)/t^2, t<1/261.217.35.142 06/05 15:58

9^F→suhorng:∫(0,1/2] dt/t^2 →∞61.217.35.142 06/05 15:59

10^F→suhorng:∞-∞可以想辦法弄成0/0或∞/∞ 通分之類61.217.35.142 06/05 16:21

11^F→suhorng:抱歉第四行應是lim sin(u)/u = 161.217.35.142 06/05 16:22

15^F→suhorng:請驗證∫[0,x]sin(1/(1+t))dt以及∫[x,1]61.217.35.142 06/05 20:55

16^F→suhorng:cos(t)/t^2 dt 都發散所以第一題是∞/∞61.217.35.142 06/05 20:55

17^F→suhorng:第三題是0．∞ 都可以試試看l'Hopital61.217.35.142 06/05 20:55

[多變] 95下台大微甲一組期中

[ trans_math ]3 留言, 推噓總分: +1

作者: jjx66os - 發表於 2012/06/02 23:32(12年前)

1^F推suhorng:Gx=0:x(3x-2)+3y(3x-2)=0→(x+3y)(3x-2)=061.217.34.17 06/02 23:57

2^F→suhorng:1.x=2/3代到Gy=0求y 2.x=-3y代到Gy=0求x,y61.217.34.17 06/02 23:58

[多變]92下台大微乙期中

[ trans_math ]10 留言, 推噓總分: 0

作者: jjx66os - 發表於 2012/06/02 16:24(12年前)

1^F→suhorng:cos(x) + cos(x+y) = 2cos(x+y/2)cos(y/2)61.217.34.17 06/02 21:48

2^F→suhorng:cos(y) + cos(x+y) = 2cos(x/2+y)cos(x/2)61.217.34.17 06/02 21:48

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