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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: allstar on August 05, 2014, 04:51:20 pm

Title: Antidifferentiation Question
Post by: allstar on August 05, 2014, 04:51:20 pm: how would i antidifferentiate [x^4+12x^3+36x^2]^-1?

Moderator Action: Changed topic name from "hello?" to something more relevant.
Title: Re: hello?
Post by: keltingmeith on August 05, 2014, 05:23:44 pm: Factorise the denominator and split it up into partial fractions.
Title: Re: hello?
Post by: allstar on August 05, 2014, 06:15:05 pm: when i factorise i get x^2 (x+6)^2

A(x+6)^2 + Bx^2 = 1

if x = 0 A = 1/36
if x = -6 B = 1/36

so the antidiff is 1/36 integral( 1/x^2 + 1/(x+6)^2)
1/36 [-x^-1 - (x+6)^-1 ]
Title: Re: Antidifferentiation Question
Post by: keltingmeith on August 05, 2014, 06:47:47 pm: Just tag on a constant of integration, and it looks fine to me.
Title: Re: Antidifferentiation Question
Post by: allstar on August 06, 2014, 11:06:45 am: thanks

i have a euler's question:

its from the 2011 vcaa exam 2

why did they assume that h is equal to the dy/dx multiplied by t?
Title: Re: Antidifferentiation Question
Post by: polar1 on August 06, 2014, 02:40:45 pm: Quote from: allstar on August 06, 2014, 11:06:45 am
thanks

i have a euler's question:

its from the 2011 vcaa exam 2

why did they assume that h is equal to the dy/dx multiplied by t?

I just looked at the examiner's report comment for that question, and they did it the 'standard way'. $h$ corresponds to the step size anyway, which, in this case is 1/2.