ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kenhung123 on June 29, 2010, 05:48:25 pm

Title: Integration Cases
Post by: kenhung123 on June 29, 2010, 05:48:25 pm: Do we need to know any integration cases other than $(ax+b)^{n}$ , $\frac{1}{ax+b}$ , $e^{kx}$ , $sin(kx+a)$ and $cos(kx+a)$ ?

Personally I learnt $sin^{a}(x)cos^{b}(x)$ just in case but wondering would it be better if a I learn more for other functions also.

Thanks in advance
Title: Re: Integration Cases
Post by: brightsky on June 29, 2010, 06:03:06 pm: Nup, don't think so. But it's always good/fun to learn a few more cases. :p
Title: Re: Integration Cases
Post by: Ilovemathsmeth on July 01, 2010, 09:38:05 pm: Nope, no more than what you said above. However I do advise practicing integration by recognition as that is a very useful technique.

brightsky, I agree with you, other cases are fun :)
Title: Re: Integration Cases
Post by: TyErd on July 01, 2010, 10:01:49 pm: Whats integration by recognition?
Title: Re: Integration Cases
Post by: Ilovemathsmeth on July 01, 2010, 11:30:22 pm: That's when you are asked to differentiate a function and then find an antiderivative for something that looks very similar to the original function, maybe there is a coefficient difference or a constant difference or some other function difference that is easy to integrate.

E.g. differentiate xsin(x)

That gives: xcos(x) + sin(x).

Then find an antiderivative for 2xcos(x) + 2sin(x).
Title: Re: Integration Cases
Post by: TyErd on July 01, 2010, 11:51:31 pm: Ohh yeah i did a couple of those today, thnx for clearing that up
Title: Re: Integration Cases
Post by: m@tty on July 02, 2010, 12:12:56 am: Or, find the derivative of $x\log_e{x}$ and hence anti differentiate $\log_e{x}$ .

That is a nice one to remember.
Title: Re: Integration Cases
Post by: TyErd on July 02, 2010, 11:30:19 am: is the answer to the above problem this:

$\frac {dy}{dx} = 1+log_e (x) <br /><br /> and \int log_e (x) dx = x log_e (x) - \int 1 + c$
Title: Re: Integration Cases
Post by: the.watchman on July 02, 2010, 11:35:18 am: Quote from: TyErd on July 02, 2010, 11:30:19 am
is the answer to the above problem this:

$\frac {dy}{dx} = 1+log_e (x) <br /><br /> and \int log_e (x) dx = x log_e (x) - \int 1 + c$

Yes, but you should write it as $xlog_e (x) - x + c$ :P
Title: Re: Integration Cases
Post by: kenhung123 on July 02, 2010, 11:35:37 am: yea but you can simplify the integral of 1 to x
Title: Re: Integration Cases
Post by: TrueTears on July 02, 2010, 11:39:46 am: Quote from: TyErd on July 01, 2010, 10:01:49 pm
Whats integration by recognition?
basically integration by parts in disguise.
Title: Re: Integration Cases
Post by: TyErd on July 02, 2010, 12:04:21 pm: okay thanks for that
Title: Re: Integration Cases
Post by: kenhung123 on July 02, 2010, 12:12:02 pm: lol is that sarcasm? :P
Title: Re: Integration Cases
Post by: the.watchman on July 02, 2010, 12:13:32 pm: Quote from: TyErd on July 02, 2010, 12:04:21 pm
okay thanks for that

Oh, and $\int 1dx$ , my friend :P
Title: Re: Integration Cases
Post by: TyErd on July 02, 2010, 12:19:12 pm: oh yeaah I always forget that thankyou! Nah wasn't sarcasm kenhung123 lol
Title: Re: Integration Cases
Post by: Ilovemathsmeth on July 02, 2010, 07:02:49 pm: never ever forget the 'dx'. Please... :P
Title: Re: Integration Cases
Post by: Whatlol on July 03, 2010, 11:08:38 pm: Quote from: Ilovemathsmeth on July 02, 2010, 07:02:49 pm
never ever forget the 'dx'. Please... :P

i dont like putting dx.
Title: Re: Integration Cases
Post by: superflya on July 03, 2010, 11:17:20 pm: its not the dx's its the effing +c that i dont like.
Title: Re: Integration Cases
Post by: Whatlol on July 03, 2010, 11:44:45 pm: Quote from: superflya on July 03, 2010, 11:17:20 pm
its not the dx's its the effing +c that i dont like.
yea thats worse i forgot it everytime!
Title: Re: Integration Cases
Post by: Cthulhu on July 03, 2010, 11:54:25 pm: Quote from: superflya on July 03, 2010, 11:17:20 pm
its not the dx's its the effing +c that i dont like.
I always lost a mark or two by forgetting +c's
Title: Re: Integration Cases
Post by: Blakhitman on July 04, 2010, 02:10:44 am: Quote from: superflya on July 03, 2010, 11:17:20 pm
its not the dx's its the effing +c that i dont like.

THIS.
Title: Re: Integration Cases
Post by: kenhung123 on July 14, 2010, 06:45:53 pm: Hmm, I'm confused, I read somewhere when they ask find 'an' antiderivative we don't need +C but in essentials Ex 12C Q2, the answers includes +C. So I'm actually quite worried about this wording thing. To confirm "find the RULE/THE antiderivative" we need +C?
Title: Re: Integration Cases
Post by: Blakhitman on July 14, 2010, 07:03:07 pm: Yes.
Title: Re: Integration Cases
Post by: kenhung123 on July 14, 2010, 07:08:06 pm: What about in the case of finding 'an' antiderivative?
Title: Re: Integration Cases
Post by: TrueTears on July 14, 2010, 07:09:13 pm: an means a class of functions, no need for +c

the is a specific function of that class, need for +c
Title: Re: Integration Cases
Post by: kenhung123 on July 14, 2010, 07:10:16 pm: Quote from: TrueTears on July 14, 2010, 07:09:13 pm
an means a class of functions, no need for +c

the is a specific function of that class, need for +c
But umm, I am a bit confused because Ex12C q2 includes +C in the case of 'an antiderivative'
Title: Re: Integration Cases
Post by: Blakhitman on July 14, 2010, 07:24:14 pm: Quote from: TrueTears on July 14, 2010, 07:09:13 pm
an means a class of functions, no need for +c

the is a specific function of that class, need for +c

I've always wondered though, what if you put some arbitrary number as a constant?
Title: Re: Integration Cases
Post by: m@tty on July 14, 2010, 07:41:40 pm: It would be acceptable. Just some explanation may be required.
Title: Re: Integration Cases
Post by: TrueTears on July 14, 2010, 07:43:25 pm: Quote from: Blakhitman on July 14, 2010, 07:24:14 pm
Quote from: TrueTears on July 14, 2010, 07:09:13 pm
an means a class of functions, no need for +c

the is a specific function of that class, need for +c

I've always wondered though, what if you put some arbitrary number as a constant?
in the eyes of a mathematician thatd be fine, heck most mathematicians im sure probs wont even bother with +c if it was a trivial "integrate this" question, however some tightass vce examiner may think not.
Title: Re: Integration Cases
Post by: m@tty on July 14, 2010, 07:48:39 pm: Quote from: kenhung123 on July 14, 2010, 07:10:16 pm
Quote from: TrueTears on July 14, 2010, 07:09:13 pm
an means a class of functions, no need for +c

the is a specific function of that class, need for +c
But umm, I am a bit confused because Ex12C q2 includes +C in the case of 'an antiderivative'
If you want to be absolutely sure, check out Q 2.a. from exam 1 last year. VCAA left out the '+C'
Title: Re: Integration Cases
Post by: kenhung123 on July 14, 2010, 09:18:07 pm: Why is it that $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|2x-5|+C$ not $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|5-2x|+C$ but $\int \frac{1}{(5-2x)^{2}}dx=\frac{1}{2}(5-2x)^{-1}+C$
Title: Re: Integration Cases
Post by: tcg93 on July 14, 2010, 09:19:43 pm: Quote from: kenhung123 on July 14, 2010, 09:18:07 pm
Why is it that $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|2x-5|+C$ not $\int \frac{1}{5-2x}dx=\frac{-1}{2}log_e|5-2x|+C$ but $\int \frac{1}{(5-2x)^{2}}dx=\frac{1}{2}(5-2x)^{-1}+C$

The mod makes them the same thing - try subbing any value of x in - you will get the same result. 2x - 5 just looks nicer.
Title: Re: Integration Cases
Post by: kenhung123 on July 15, 2010, 10:17:48 pm: thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?
Title: Re: Integration Cases
Post by: TrueTears on July 15, 2010, 10:18:38 pm: Quote from: kenhung123 on July 15, 2010, 10:17:48 pm
thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?
differentiate the RHS and you will see why :D
Title: Re: Integration Cases
Post by: kenhung123 on July 15, 2010, 10:35:20 pm: Yea but why doesn't that formula apply? (ax+b)^-1
Title: Re: Integration Cases
Post by: TrueTears on July 15, 2010, 10:36:35 pm: i dono that formula, i just do it manually.
Title: Re: Integration Cases
Post by: Martoman on July 15, 2010, 11:00:31 pm: Quote from: kenhung123 on July 15, 2010, 10:17:48 pm
thanks

why is it that $\int \frac{1}{2x}dx\neq \frac{1}{x}log_e(2x)+C$ ?

Ok this is really.basic. but take out 1/2 then you have 1/x -> .5ln|x|
Title: Re: Integration Cases
Post by: tcg93 on July 16, 2010, 12:56:34 pm: Quote from: kenhung123 on July 15, 2010, 10:35:20 pm
Yea but why doesn't that formula apply? (ax+b)^-1

yeah don't use the formula because you may "forget" it in the exam, just manipulate the integral moving coefficients outside the integral so you get a function's derivative on top of the function
Title: Re: Integration Cases
Post by: Martoman on July 16, 2010, 02:19:25 pm: Quote from: kenhung123 on July 15, 2010, 10:35:20 pm
Yea but why doesn't that formula apply? (ax+b)^-1

it doesn't apply because $(ax+b)^{-1+1} = (ax+b)^0 = 1$ , a trivial case where the power rule fails.