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Author Topic: Antidifferentiation By Deduction  (Read 2406 times)  Share 

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squance

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Antidifferentiation By Deduction
« on: October 30, 2007, 10:38:11 am »
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Can someone please give me an example of how this works??? What's the likelihood of this being in the exam?

cara.mel

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Antidifferentiation By Deduction
« Reply #1 on: October 30, 2007, 10:50:26 am »
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Do you mean like:
a) Differentiate this
b) Hence, antidifferentiate this?
or something else

reg

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Antidifferentiation By Deduction
« Reply #2 on: October 30, 2007, 11:17:33 am »
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Integration by recognition is what my textbook calls it. IMO it (can be/is) the hardest thing on the course.

Collin Li

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Antidifferentiation By Deduction
« Reply #3 on: October 30, 2007, 11:49:55 am »
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Quote from: "reg"
Integration by recognition is what my textbook calls it. IMO it (can be/is) the hardest thing on the course.


It is probably one of the easiest things to do. There is no thinking required, just follow the steps that VCAA hand out for you. It's also a matter of setting out your notation well:

Example

a. Find the derivative of xlogx with respect to x.
d[xlogx]/dx = (1)*logx + x*(1/x) = logx + 1
(I strongly recommend this notation because it will be clear how to proceed in the next step)

b. Hence, find the antiderivative of logx.
Integrate both sides with respect to x:
=> xlogx = INTEGRATE[logx] + x
=> INTEGRATE[logx] = xlogx - x
(The anti-derivative cancels out with the derivative)

That's what you yield algebraically. You should however, note that there is a constant of integration within that integral. Your general solution would be:

INTEGRATE[logx] = xlogx - x + C

edited to correct for a stupid mistake; thanks Ahmad

Ahmad

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« Reply #4 on: October 30, 2007, 11:56:53 am »
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Check your working carefully!  8)
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asa.hoshi

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« Reply #5 on: October 30, 2007, 12:46:05 pm »
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prefect exam 1 question i reckon.
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Eriny

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« Reply #6 on: October 30, 2007, 02:46:27 pm »
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I actually like integration by recognition. I can do it! Besides, they practically give you the answer anyway.

Collin Li

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Antidifferentiation By Deduction
« Reply #7 on: October 30, 2007, 05:15:09 pm »
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Quote from: "Ahmad"
Check your working carefully!  8)


Thanks, I should know by now that whenever you end up with a numerically defined constant from integrating, you've done something wrong, lol.

2happy

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« Reply #8 on: October 30, 2007, 08:19:03 pm »
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i dont think it''ll be on the exam1
e: 16 year old against them: 17-18 year olds

NO EXAMS left this year

Spesh, Kehm, Fysiks, Ing, C SL 2008


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melanie.dee

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« Reply #9 on: October 30, 2007, 08:23:58 pm »
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how come? im thinking it'll definitely be there. and i hate it! along with the entire rest of the methods course. although im strangely attracted to doing a bit of study for methods at the moment.. must be that looming english exam haha. trying to put off study for that 8)

2happy

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« Reply #10 on: October 30, 2007, 08:29:37 pm »
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theyve got other more important things to test
e: 16 year old against them: 17-18 year olds

NO EXAMS left this year

Spesh, Kehm, Fysiks, Ing, C SL 2008


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Toothpaste

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Antidifferentiation By Deduction
« Reply #11 on: October 30, 2007, 08:52:58 pm »
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No they don't.

It's a likely question.

2happy

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« Reply #12 on: October 30, 2007, 08:56:54 pm »
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exam 2 mcq
e: 16 year old against them: 17-18 year olds

NO EXAMS left this year

Spesh, Kehm, Fysiks, Ing, C SL 2008


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Mix it Up TM

asa.hoshi

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Antidifferentiation By Deduction
« Reply #13 on: October 30, 2007, 10:35:56 pm »
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i believe it is a likely question in exam 1. because they didn't test it last year. this question can test student's differential and anti-differential skills/techniques. if its not on exam 1, it will be on exam 2 FOR SURE (i think)!

they're probably fit in a related-rates question in exam 1 as well since it wasn't in last year's paper.
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