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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: bucket on February 02, 2008, 11:53:46 pm

Title: Integration
Post by: bucket on February 02, 2008, 11:53:46 pm: I'm pretty sure this question is pretty basic, but I'm stupid and don't even know/forgot how to find derivatives LOL.

Evaluate:

$\displaystyle\int^5_1 3 - x\, dx$
Title: Re: Integration
Post by: Collin Li on February 02, 2008, 11:56:48 pm: $\displaystyle\int^5_1 (3 - x)\, dx = \left[ 3x-\frac{1}{2}x^2 \right]^5_1 = \left(15 - \frac{25}{2}\right) - \left(3 - \frac{1}{2}\right) = 0$

This is just substituting in values after you recognise that the anti-derivative of $3-x$ is $3x-\frac{1}{2}x^2$ (plus a constant).
Title: Re: Integration
Post by: AppleXY on February 02, 2008, 11:57:55 pm: are you serious ? :P

wow. I'm extremely bad at teaching but hey :P

$[3x - \frac{x^2}{2}]$ 1 5

$[3(5) - \frac{(5)^2}{2}]$ $-$ $[3(1) - \frac{(1)^2}{2}]$

Evaluate.

Fuck someone bet me :P
Title: Re: Integration
Post by: bucket on February 03, 2008, 12:07:41 am: lol how embarrassing.
thanks
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 12:11:02 am: Quote from: bucket on February 03, 2008, 12:07:41 am
lol how embarrassing.
thanks

Also, you (and I) should have written the expression as $\displaystyle\int^5_1 (3 - x)\, dx$

Even though the "dx" makes it clear where the expression ends, you could still get marked down by an anal examiner for lacking the brackets around the expression (since it's an integration over more than one term). Something like $\displaystyle\int^5_1 x^2\sin x\, dx$ doesn't need brackets because it's just one term (even though it's a lot more complicated!)
Title: Re: Integration
Post by: AppleXY on February 03, 2008, 12:11:37 am: lol. No question is embrassing. Feel free to ask anything (even $1+1$ , ok not that but you get the idea). Collin and Ahmad will answer it professionally :D

in general:

(http://img184.imageshack.us/img184/7173/41aa2fdb1f9f344e4e991cfce9.png)

where F(a) and F(b) are the integrated functions (the original ones)
Title: Re: Integration
Post by: costargh on February 03, 2008, 12:14:00 am: Whats an example of a hard integration? For the Methods course?
Title: Re: Integration
Post by: dcc on February 03, 2008, 12:18:02 am: $\int e^{x^2} dx$

is one ive seen on exam papers before
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 12:19:26 am: Quote from: costargh on February 03, 2008, 12:14:00 am
Whats an example of a hard integration? For the Methods course?

The toughest ones you can get are going to be things like $\displaystyle\int^5_1 (2x+3)^6\, dx$ (not very hard!). There are also tricky ones where you have to use recognition, but they are actually not hard once you've seen it and done it a few times.

Here's an example (the technique is called integration by parts, but you don't need to know about this):

a) Find $\frac{d}{dx}\left(xe^x\right)$
b) Hence, find $\displaystyle\int xe^x\, dx$

BTW, dcc is an idiot, that integral can't be done in terms of elementary functions.
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 12:24:35 am: Quote from: coblin on February 03, 2008, 12:19:26 am
a) Find $\frac{d}{dx}\left(xe^x\right)$
b) Hence, find $\displaystyle\int xe^x\, dx$

The way these problems are solved:

$\frac{d}{dx}\left(xe^x\right) = e^x + xe^x$

Integrating both sides with respect to x yields:
$xe^x = \displaystyle\int \left(e^x + xe^x\right)\, dx$

$\implies xe^x = \displaystyle\int e^x\, dx + \displaystyle\int xe^x\, dx$

$\implies xe^x - \displaystyle\int e^x\, dx = \displaystyle\int xe^x\, dx$

$\therefore \displaystyle\int xe^x\, dx = xe^x - e^x + C$

You don't have to think in these problems much, you just do what you're told in the first part, then you integrate both sides and you rearrange to get the integral asked in the second part.
Title: Re: Integration
Post by: costargh on February 03, 2008, 12:25:19 am: Quote from: coblin on February 03, 2008, 12:19:26 am
BTW, dcc is an idiot, that integral can't be done in terms of elementary functions.

LOL. I dont even know what that means but LOL.

But I have seen harder ones then that coblin. Like ones with e. dont they become log e or someting?
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 12:27:39 am: You mean like $\displaystyle\int \frac{1}{x}\, dx = \log_ex + C$ ?

These are quite easy. Any linear polynomial taken to the power of -1 integrates into the natural logarithm. (It's just a special case of the integration of a linear polynomial taken to the $n$ th power - normally you add one to the power and divide by it, but for $n=-1$ , you can't because you'd be dividing by zero)

Come to think of it, there was a quite good integration question in 2005 which involved integrating a power function that wasn't base $e$ . I'll have a look at it in my spare time, but you basically had to convert the base into $e$ so that you could work with it easily (because we know the integral of the exponential function). Basically, we know how to integrate $e^x$ , but we don't know how to integrate $2^x$ , but can we manipulate the latter into a similar form to the exponential?
Title: Re: Integration
Post by: AppleXY on February 03, 2008, 12:32:15 am: Quote from: Coblin
$\displaystyle\int^5_1 (2x+3)^6\, dx$

I'd use integration by substitution rather than expanding it.
Title: Re: Integration
Post by: midas_touch on February 03, 2008, 12:32:33 am: Quote from: dcc on February 03, 2008, 12:18:02 am
$\int e^{x^2} dx$

is one ive seen on exam papers before

Coblin, didnt we do problems like these at uni last year where u have to use series to solve such integrals?
Title: Re: Integration
Post by: midas_touch on February 03, 2008, 12:33:20 am: Quote from: AppleXY on February 03, 2008, 12:32:15 am
Quote from: Coblin
$\displaystyle\int^5_1 (2x+3)^6\, dx$

I'd use integration by substitution rather than expanding it.

Whoever tries to expand this before integrating is a noobcake.
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 12:41:55 am: Quote from: midas_touch on February 03, 2008, 12:32:33 am
Quote from: dcc on February 03, 2008, 12:18:02 am
$\int e^{x^2} dx$

is one ive seen on exam papers before

Coblin, didnt we do problems like these at uni last year where u have to use series to solve such integrals?

Yeah, but that's not elementary functions.

Quote from: AppleXY on February 03, 2008, 12:32:15 am
Quote from: Coblin
$\displaystyle\int^5_1 (2x+3)^6\, dx$

I'd use integration by substitution rather than expanding it.

Yeah, you're not supposed to expand it. Since it's a linear function on the inside, you can make a substitution. However, in the Methods course, you are not taught this, so you have to stick to the cumbersome formula:

$\displaystyle\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)}$

This is essentially derived from substitution, but can be verified by differentiating both sides (which is how I remember it).
Title: Re: Integration
Post by: Neobeo on February 03, 2008, 12:48:09 am: $\int x^x dx$
Title: Re: Integration
Post by: AppleXY on February 03, 2008, 09:09:48 am: Hey, Collin, would you lose marks if you used the subsitution method? (I mean, without straight going to the formula? ). lol.
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 09:21:29 am: Quote from: AppleXY on February 03, 2008, 09:09:48 am
Hey, Collin, would you lose marks if you used the subsitution method? (I mean, without straight going to the formula? ). lol.

Not sure. I didn't use the substitution method. You won't need it with a bit of practice. It's a lot quicker without it. Pretty much you're doing the substitution in your head.
Title: Re: Integration
Post by: AppleXY on February 03, 2008, 10:00:41 am: lol. I know, just asking -- testing how flexible the VCAA is :P
Title: Re: Integration
Post by: bucket on February 03, 2008, 08:03:31 pm: Another probably shit easy question, wtf at the fraction =\

Find Y in terms of X if:
$\frac{dy}{dx} = 4x^{-1/2}$ and $y=1$ when $x=9$
Title: Re: Integration
Post by: Mao on February 03, 2008, 08:24:43 pm: What you have basically is a function to integrate, and a set of coordinates so you can work out $C$ .

$y = \int 4x^{-\frac{1}{2}} dx = \frac{4 \cdot x^{\frac{1}{2}}}{\frac{1}{2}} + C= 8\sqrt{x} + C$

given that the coordinates are $(9,1)$ , substituting gives:

$1 = 8\sqrt{9} + C$

$C=-23$ / $C=25$

however, if $C=25$ , it'd imply that the graph's gradient is negative, this certainly doesnt satisfy the $\frac{dy}{dx}$ that was given to us

Hence: $y=8\sqrt{x}-23$ given that $y \in [-23,\infty)$

hope that made sense :P :D
Title: Re: Integration
Post by: bucket on February 03, 2008, 09:09:01 pm: Hmmm
I don't really understand the steps you take completely, and why you changed it to 23 from 25, can you just change the answer like that? lol
Sorry I'm pretty much a methods novice, trying to pick up after a year of bludging =\\
Can someone else clarify the answer?
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 09:48:56 pm: I have no idea what he is doing either.

Quote from: Mao on February 03, 2008, 08:24:43 pm
What you have basically is a function to integrate, and a set of coordinates so you can work out $C$ .

$y = \int 4x^{-\frac{1}{2}} dx = \frac{4 \cdot x^{\frac{1}{2}}}{\frac{1}{2}} + C= 8\sqrt{x} + C$

given that the coordinates are $(9,1)$ , substituting gives:

$1 = 8\sqrt{9} + C$

Up to this part makes sense. Now you find $C$ and you get:

$C = 1 - 8\sqrt{9} = 1 - 8(3) = 1 - 24 = -23$

$\therefore y = 8\sqrt{x} - 23$
Title: Re: Integration
Post by: Mao on February 03, 2008, 10:21:15 pm: $\sqrt{9}$ could also be $-3$ tho, so i thought i'd be pragmatic :P

and btw coblin i think u meant
$\therefore y = 8\sqrt{x} - 23$
in ur last line :P
Title: Re: Integration
Post by: bucket on February 03, 2008, 10:22:31 pm: Ah I get it now!
Thanks
Title: Re: Integration
Post by: Collin Li on February 03, 2008, 10:34:54 pm: Quote from: Mao on February 03, 2008, 10:21:15 pm
$\sqrt{9}$ could also be $-3$ tho, so i thought i'd be pragmatic :P

No it can't. The square root sign is a positive number (for real values at least). This is why you always have to say $x^2 = a \implies x = \pm \sqrt{a}$ .

Quote
and btw coblin i think u meant
$\therefore y = 8\sqrt{x} - 23$
in ur last line :P

Thanks, fixed.