Author Topic: Integration (Read 5563 times) Tweet Share

Collin Li · « **Reply #15 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).

Neobeo · « **Reply #16 on:** February 03, 2008, 12:48:09 am »

AppleXY · « **Reply #17 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.

Collin Li · « **Reply #18 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.

AppleXY · « **Reply #19 on:** February 03, 2008, 10:00:41 am »

lol. I know, just asking -- testing how flexible the VCAA is

bucket · « **Reply #20 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 $x=9$

Mao · « **Reply #21 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

bucket · « **Reply #22 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?

Collin Li · « **Reply #23 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:

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

$\therefore y = 8\sqrt{x} - 23$

Mao · « **Reply #24 on:** February 03, 2008, 10:21:15 pm »

$\sqrt{9}$ could also be $-3$ tho, so i thought i'd be pragmatic

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

bucket · « **Reply #25 on:** February 03, 2008, 10:22:31 pm »

Ah I get it now!
Thanks

Collin Li · « **Reply #26 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

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

Thanks, fixed.

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