ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Yoda on October 18, 2014, 08:49:26 pm

Title: recognition question help?
Post by: Yoda on October 18, 2014, 08:49:26 pm: given that the derivative of 5/2x^2-1 is -20x/(2x^2-10^2. Find the value of a given that integral (upper limit a, lower limit 1) of
(20x/(2x^2-1)^2 +1)dx =37/7.
Title: Re: recognition question help?
Post by: Avainer on October 25, 2014, 01:14:17 pm: This is so hard to read, you should work on formating and use more brackets. In working and on forum. Assuming I read it right, you need to find integral(low 1 top a) of the second eqn given and put that as equal to 37/3. Then, the thing you put in the square brackets is the first equation given. You need up with [5/(2xs^2 -1)](low 1 top a) = 37/3 Then solve for a. Hope that was helpful.
Title: Re: recognition question help?
Post by: myanacondadont on October 25, 2014, 05:23:49 pm: Can you not just; $\int_{1}^{a}\frac{20x}{(2x^2-1)^2}+1.dx = \frac{37}{7}$
Put it in calc and solve for a? I got 1.356.

It sounds like an integration by recognition problem but I can't quite make it out so I just did that.
Title: Re: recognition question help?
Post by: keltingmeith on October 25, 2014, 07:46:25 pm: Quote from: myanacondadont on October 25, 2014, 05:23:49 pm
Can you not just; $\int_{1}^{a}\frac{20x}{(2x^2-1)^2}+1.dx = \frac{37}{7}$
Put it in calc and solve for a? I got 1.356.

It sounds like an integration by recognition problem but I can't quite make it out so I just did that.

VCAA will want an exact value - you've given a numerical value.
Not to mention if you need to show method or this is exam 1, you're in trouble. You should be able to do any question you see both with and without a calculator.
Title: Re: recognition question help?
Post by: Yoda on October 26, 2014, 06:44:51 pm: Its tech free