Author Topic: Integration of factorised functions by hand? (Read 2080 times) Tweet Share

Duck · « **on:** September 26, 2010, 04:06:29 pm »

These two questions have come up on Kilbaha and Itute exam 1's so it's fair to say that they're at the higher end of questions but I can't figure them out:
Integrate (2x-1)^3

Integrate 4/(2x+3)^2

Anyone know how one would go about doing this without a calculator?

taiga · « **Reply #1 on:** September 26, 2010, 04:11:29 pm »

There is a general form for this which is not on the formula sheet, I don't know how to use latex, but I'll try find you an image.

If you use essentials mathematics it should be there in the form (ax+b)^n

Duck · « **Reply #2 on:** September 26, 2010, 04:15:32 pm »

Yeh thanks I just looked in the textbook and found the chapter on it. Why the hell would they include every formula for integration but that one on the data sheet?

taiga · « **Reply #3 on:** September 26, 2010, 04:17:37 pm »

Quote from: Duck on September 26, 2010, 04:15:32 pm

Yeh thanks I just looked in the textbook and found the chapter on it. Why the hell would they include every formula for integration but that one on the data sheet?

I don't know, it took me quite a while to understand those questions until i finally saw that =\

98.40_for_sure · « **Reply #4 on:** September 26, 2010, 04:17:56 pm »

the.watchman · « **Reply #5 on:** September 26, 2010, 04:42:37 pm »

Quote from: 99.95_for_sure on September 26, 2010, 04:17:56 pm

$\int (2x-1)^{3}dx$

$=\frac{(2x-1)^{4}}{4*2}=\frac{(2x-1)^{4}}{8}$

$\int 4(2x+3)^{-2}dx$

$=\frac{4(2x+3)^{-1}}{-1*2}$

$=\frac{-2}{2x+3}$

NOOOOOO!!! PLUS 'C'

8039 · « **Reply #6 on:** September 26, 2010, 04:43:46 pm »

Quote from: 99.95_for_sure on September 26, 2010, 04:17:56 pm

$\int (2x-1)^{3}dx$

$=\frac{(2x-1)^{4}}{4*2}=\frac{(2x-1)^{4}}{8}$

$\int 4(2x+3)^{-2}dx$

$=\frac{4(2x+3)^{-1}}{-1*2}$

$=\frac{-2}{2x+3}$

What if it were something like (3x^2 - 2x)^-2?

Duck · « **Reply #7 on:** September 26, 2010, 04:45:11 pm »

Haha watchman. I hate that so much. Even worse was when I lost a mark because I put +c in when it was asking for 'an' antiderivative.

the.watchman · « **Reply #8 on:** September 26, 2010, 04:47:09 pm »

Quote from: Duck on September 26, 2010, 04:45:11 pm

Haha watchman. I hate that so much. Even worse was when I lost a mark because I put +c in when it was asking for 'an' antiderivative.

Lol, whenever they ask for an antiderivative, I write:

$\int f(x)dx = ... + c$

So an antiderivative is $...$

Duck · « **Reply #9 on:** September 26, 2010, 04:48:47 pm »

When they ask for 'THE' antiderivative you have to put +c but when they ask for 'AN' antiderivative you don't put +c.

stonecold · « **Reply #10 on:** September 26, 2010, 04:50:26 pm »

Quote from: 8039 on September 26, 2010, 04:43:46 pm

Quote from: 99.95_for_sure on September 26, 2010, 04:17:56 pm
$\int (2x-1)^{3}dx$

$=\frac{(2x-1)^{4}}{4*2}=\frac{(2x-1)^{4}}{8}$

$\int 4(2x+3)^{-2}dx$

$=\frac{4(2x+3)^{-1}}{-1*2}$

$=\frac{-2}{2x+3}$

What if it were something like (3x^2 - 2x)^-2?

According to DHA, you can't anti diff those. At least not in methods anyway!

Edit: Well you could expand that one seeing as it is not so big, but for a larger power, I don't think we have to know it

the.watchman · « **Reply #11 on:** September 26, 2010, 04:51:47 pm »

Quote from: Duck on September 26, 2010, 04:48:47 pm

When they ask for 'THE' antiderivative you have to put +c but when they ask for 'AN' antiderivative you don't put +c.

But, when they ask for an antiderivative, if you write $\int f(x)dx = g(x)$ without the 'c', that is technically wrong.
It is more 'correct' to write the above statement with a 'c' and to follow that up with a statement:
"So an antiderivative is g(x)+1000"

naved_s9994 · « **Reply #12 on:** September 26, 2010, 05:01:42 pm »

Quote from: stonecold on September 26, 2010, 04:50:26 pm

Quote from: 8039 on September 26, 2010, 04:43:46 pm
Quote from: 99.95_for_sure on September 26, 2010, 04:17:56 pm
$\int (2x-1)^{3}dx$

$=\frac{(2x-1)^{4}}{4*2}=\frac{(2x-1)^{4}}{8}$

$\int 4(2x+3)^{-2}dx$

$=\frac{4(2x+3)^{-1}}{-1*2}$

$=\frac{-2}{2x+3}$

What if it were something like (3x^2 - 2x)^-2?

According to DHA, you can't anti diff those. At least not in methods anyway!

Edit: Well you could expand that one seeing as it is not so big, but for a larger power, I don't think we have to know it

I think you can... I've seen stuff like that on trials?
Although yeah ofcourse you cannot trust them.

m@tty · « **Reply #13 on:** September 26, 2010, 05:14:08 pm »

Okay, evaluate $\int \frac{1}{9x^4-6x^3+4x^2} dx$ gogo...

EDIT: The other integrals mentioned here can be done by the general rule given in textbooks. This works as long as the interior function is LINEAR.

The rule is just one case of integration by substitution. Which is specialist material.

Jdog · « **Reply #14 on:** September 26, 2010, 06:25:36 pm »

just put plus a +c or some sorta constant no matter what.

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Author Topic: Integration of factorised functions by hand? (Read 2080 times) Tweet Share

Duck

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