ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Maz on December 06, 2015, 08:37:45 pm

Title: integration help plz
Post by: Maz on December 06, 2015, 08:37:45 pm: hey, could someone please help me do this question?
i have absolutely no idea how to do the integration of this:
(2x+1)^4 and 12/(1-3x)^2
i really don't understand the book when it explains it...

thankyou so much in advance :)
Title: Re: integration help plz
Post by: TheAspiringDoc on December 06, 2015, 09:54:46 pm: this might help: https://au.answers.yahoo.com/question/index?qid=20110419071708AAmnMUs
Title: Re: integration help plz
Post by: cosine on December 06, 2015, 10:05:17 pm: Quote from: mq123 on December 06, 2015, 08:37:45 pm
hey, could someone please help me do this question?
i have absolutely no idea how to do the integration of this:
(2x+1)^4 and 12/(1-3x)^2
i really don't understand the book when it explains it...

thankyou so much in advance :)

There is a rule for integration when you have linear expressions raised to any power, except for 1, that is:

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

So in your example, a = 2, b = 1 and n = 4:

$\int (2x+1)^4 dx$ = $\frac{(2x+1)^5}{2(5)} + c = \frac{(2x+1)^5}{10} + c$

For the second example, you can use the same rule but you must raise the expression so that it remains not in fraction form, and you do this by negating the power:

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

Now, your a = -3, b = 1 and n = -2 and plug these into the rule I stated above:

$\int (1-3x)^{-2}dx = \frac{(1-3x)^{-1}}{-3(-1)} + c = \frac{1}{3(1-3x)} + c$

Hope this helps you a bit, let me know if you need further clarifications!
Title: Re: integration help plz
Post by: TheAspiringDoc on December 06, 2015, 10:10:03 pm: May I ask how you know that rule cosine?
Do you simply have to remember it or is there an intuitive method that you used?
Title: Re: integration help plz
Post by: Splash-Tackle-Flail on December 06, 2015, 10:31:38 pm: I'm not cosine, but the way I think of it is that in normal polynomial integration, we know that

$\int{x^n}dx=\frac{x^{n+1}}{n+1} + c$

The second step uses the property in differentiation of the chain rule.

So in integration you would divide by $\frac{du}{dx}$ in a sort of reverse chain rule, giving the rule cosine wrote.

If you want to do your own research or look for better examples, google u substitution (which is rule is a form of), which you do in spesh anyway :)
Title: Re: integration help plz
Post by: wyzard on December 07, 2015, 01:01:12 pm: Quote from: Splash-Tackle-Flail on December 06, 2015, 10:31:38 pm
I'm not cosine, but the way I think of it is that in normal polynomial integration, we know that

$\int{x^n}dx=\frac{x^{n+1}}{n+1} + c$

The second step uses the property in differentiation of the chain rule.

So in integration you would divide by $\frac{du}{dx}$ in a sort of reverse chain rule, giving the rule cosine wrote.

If you want to do your own research or look for better examples, google u substitution (which is rule is a form of), which you do in spesh anyway :)

The substitution method is sure really handy in integration 8) Unfortunately it is not taught in methods but only in specialist maths :(
Title: Re: integration help plz
Post by: Maz on December 07, 2015, 04:55:17 pm: thanks guys :)