查看完整版本: [數統] 積分問題

[數統] 積分問題

以下呢條題目點in??? 請幫忙~~謝謝=]
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ans係呢個,,,唔知點解多左個2
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[[i] 本帖最後由 pineappleEEE 於 2010-7-29 04:10 PM 編輯 [/i]]

LET U=LN(T+2)
DU=(1/T+2)DT
之後自己做啦，應該OK

[quote]原帖由 [i]0310672[/i] 於 2010-7-29 04:10 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269284&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]
LET U=LN(T+2)
DU=(1/T+2)DT
之後自己做啦，應該OK [/quote]
THX~~原個要甘做先睇到個2:L

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Use method of substitution
S[ln (t+2)]dt/(t+2)
Let u=ln (t+2)
du/dt=1/(t+2)
du=dt/(t+2)
dt=(t+2)du

S[ln(t+2)]dt/(t+2)
=S[ln u](t+2)du/(t+2)
=S(ln u)du
=(ln u)^2/2+C
=(ln (t+2))^2/2+C

其實唔一定要用substitution 有方法可以仲快,不過就唔係題題用到 你想唔想知 ?em26

回覆 #6 chanyin 的帖子

咁係咩????我好想知.....

[quote]原帖由 [i]021044[/i] 於 2010-7-29 04:55 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269391&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]
咁係咩????我好想知..... [/quote]
你介唔介意今晚先講 ,我而家唔太得閒 em46em46em46

好!!!!!

d /dt [u]ln(t+2) [/u]
         (t+2)

       [u] ln(t+2)[/u]
=         u

=  [u]   ln (u-2) [/u]
           u




sub u = t + 2
     du = dt


居然唔識做...不得不放棄M&S

既然t+2唔得
咁即係得返ln (t+2) lor......

回覆 #10 龍一刀 的帖子

如果sub u=t+2
int. lnu/u d(u-2)
=int lnu/u du
=int. lnu dlnu
=lnu]^2/2

[quote]原帖由 [i]021044[/i] 於 2010-7-29 05:21 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269487&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]
既然t+2唔得
咁即係得返ln (t+2) lor...... [/quote]


∫     [u]ln(t+2)    [/u] dt
         (t+2)

=  ∫   [u]  u    [/u]  * [color=Red](t+2)  [/color]  du
       [color=Red](t+2)[/color]

= ∫   u   du

= u^2/2

= [u][ln(t+2)][/u]  + C
        2



note :
sub u= ln (t+2)
      du = 1/(t+2) dt
(1+2)du =dt



我覺得紅色唔應該寫出黎....似是表達有點問題


THX

[quote]原帖由 [i]hinokari[/i] 於 2010-7-29 05:23 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269494&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]
如果sub u=t+2
int. lnu/u d(u-2)
=int lnu/u du
=int. lnu dlnu
=lnu]^2/2 [/quote]

∫     [u]ln(t+2)    [/u] dt
         (t+2)

= ∫        [u]lnu    [/u] du
             u



之後唔係好明



note :
sub u = t+2
     du = dt




THX

int. (du/u)=int. d(lnu)

hinokari的做法係咁
Let u=t+2
du=dt

int [ln (t+2)]dt/(t+2)
=int [ln u]du/u
因為d(ln u)=(1/u)du
=int[ln u]d(ln u)
根據公式int x^n dx=(x^n+1)/(n+1)+c
=(ln u)^2/2+c
=[ln (t+2)]^2/2+c

[quote]原帖由 [i]hinokari[/i] 於 2010-7-29 05:34 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269523&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]
int. (du/u)=int. d(lnu) [/quote]

∫     [u]ln(t+2)    [/u] dt
         (t+2)

= ∫        [u][color=Red]lnu [/color] [/u] du
         [color=Blue]    u[/color]


姐係你意思要d 紅色  定藍色 ??


THX

[quote]原帖由 [i]龍一刀[/i] 於 2010-7-29 05:38 PM 發表 [url=http://lsforum.net/board/redirect.php?goto=findpost&pid=2269529&ptid=145959][img]http://lsforum.net/board/images/common/back.gif[/img][/url]


∫     ln(t+2)     dt
         (t+2)

= ∫        lnu   du
             u


姐係你意思要d 紅色  定藍色 ??


THX [/quote]
錯晒
問題在於處理積分公式的技巧
∫  (x^n) dx<------係x及dx時先係出現(x^n+1)/(n+1)+c

但
∫  ln u/u du
因為冇一條公式係可以處理∫ ln u/u du
但我地知道(1/u) du=d(ln u)
其實係代換積分法

用代換吧~~~
設u=ln(t+2)
  du=1/t+2 dt

之後代返入得返∫u du
搞掂

呢d我一學就識em35  老師仲教左個more efficient method :L

比心機做多d exerciseem26