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March 30, 2024, 01:34:07 am

Author Topic: Antiderivative of Log  (Read 6379 times)  Share 

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hifer

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Antiderivative of Log
« on: November 02, 2007, 07:12:27 pm »
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Hi there, do we always have to put the absolute sign when antidiffing to get logs? wat if the specified limits are positive?

bilgia

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Antiderivative of Log
« Reply #1 on: November 02, 2007, 09:02:04 pm »
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i dont think in methods...its in specialist
My Subjects:
2006 I.T Systems --> 42
2007 English --> 40
         Methods --> 41
         Spec --> 38
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ENTER: 97.35


                   



 

Ahmad

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Antiderivative of Log
« Reply #2 on: November 02, 2007, 09:12:35 pm »
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No if the terminals are positive you don't have to put it. However, it doesn't take anymore time, neither is it wrong if you do.
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

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Freitag

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Antiderivative of Log
« Reply #3 on: November 02, 2007, 10:51:21 pm »
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It's just useful for things such as evaluating areas under the x-axis, when you don't know the area will come out without the absolute value signs as negatives.
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

joechan521

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Antiderivative of Log
« Reply #4 on: November 03, 2007, 02:49:33 pm »
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i remember there was one VCAA exam 1 paper that u had to put modulus after anti into a log to be able to draw the graph

its prob in 2006 sample exam1
06 method47 chinese 2nd language advanced39
07 english39 specailist44 accounting44 further48 psychology35
07 ENTER 99.15

tankia

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Antiderivative of Log
« Reply #5 on: November 03, 2007, 02:57:42 pm »
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Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
tankAsaurus ;D

Freitag

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Antiderivative of Log
« Reply #6 on: November 03, 2007, 06:04:22 pm »
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Quote from: "tankia"
Wtf are you guys talking about *is intimidated* I don't know this stuff ><;


Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.

Ie. dy/dx = ( 1/x ) --> y= log_e |x|
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

tankia

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Antiderivative of Log
« Reply #7 on: November 04, 2007, 09:05:35 am »
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Quote from: "Freitag"
Quote from: "tankia"
Wtf are you guys talking about *is intimidated* I don't know this stuff ><;


Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.

Ie. dy/dx = ( 1/x ) --> y= log_e |x|


Whyyy? Lol ....I'm so screwed for exams ><;
tankAsaurus ;D

Odette

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Antiderivative of Log
« Reply #8 on: November 04, 2007, 09:33:26 am »
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Poor Tankia :( there there, i'm sure you'll be fine =]

Collin Li

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Antiderivative of Log
« Reply #9 on: November 04, 2007, 10:38:51 am »
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Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;


dy/dx = 1/x, find y: for R\{0}

Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.

Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.

tankia

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Antiderivative of Log
« Reply #10 on: November 04, 2007, 11:20:23 am »
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Quote from: "coblin"
Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;


dy/dx = 1/x, find y: for R\{0}

Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.

Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.


_.... You lost me at dy/dx >_>;
tankAsaurus ;D

Odette

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Antiderivative of Log
« Reply #11 on: November 04, 2007, 11:30:05 am »
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Quote from: "tankia"
Quote from: "coblin"
Quote from: "tankia"
Whyyy? Lol ....I'm so screwed for exams ><;


dy/dx = 1/x, find y: for R\{0}

Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.

Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.


_.... You lost me at dy/dx >_>;


Tanika I'll explain it to you on msn =]

tankia

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Antiderivative of Log
« Reply #12 on: November 04, 2007, 11:31:05 am »
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Okay :D
tankAsaurus ;D

Freitag

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Antiderivative of Log
« Reply #13 on: November 04, 2007, 03:58:26 pm »
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It seriously doesn't matter too much :p
rediction of Scores:

English: 40 (hopefully) ; Spec: 40, Methods: 40-45, Physics 35, Enhancement: HD. Last year- History 38; I.T 42.

Hopeful ENTER 98. (I need it for my scholarship :()

tankia

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Antiderivative of Log
« Reply #14 on: November 04, 2007, 07:45:15 pm »
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Quote from: "Freitag"
It seriously doesn't matter too much :p


LOL. I was doing my FIRST practice exam (one that i actually finished...Thank the lord for solutions :D) and i finally got what you guys were talking about...kinda...some what..maybe...maybe not...Okay...Yea..I'll go back to procrastinating :D
tankAsaurus ;D