Please help with this easy question. (1 Viewer)

[Shoom]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2008exams/pdf_doc/2008HSC-mathematics.pdf

Q6c. Please show working.

 

[sannous1]

hey i dont know how to do all the symbols and that and im jst gona do brief working out. so really what u have to do is make y= 5/(x-2) , to be y^2 = 25/(x-2)^2 From here you have to integrate y^2 = 25/(x-2)^2 which u end up getting: -25/(x-2)now just sub in 6 and 3 v = π [-25/(6-2)] - [-25/(3-2)] = 75π / 4 u^3 (18.75π )

 

[Shoom]

How is the integral of 5/(x-2)^2 -25(x-2).


Thats what im stuck at.....

 

[raniaaa]

How is the integral of 5/(x-2)^2 -25(x-2).


Thats what im stuck at.....

y^2 = 5^2 / (x-2)^2
y^2 = 25 (x-2)^-2 (bring the x-2 to the top)
intergrate that:
[ 25 (x-2)^-1 / -1]
[-25 / (x-2) ] (take the x-2 back down)

 

[sannous1]

ok y^2 = 5² / (x-2)²y^2 = 25/ (x-2)²now we integreate y^2= 25/ (x-2)²so u move up the (x-2)² which u get y^2= 25 (x-2) -² now when u integrate that, it is 25 (x-2)^-1 divide al that by -1 (because u add one when u integrate)NOW from (25 (x-2)^-1 divide al that by -1) , u move the (x-2)^-1 back down, and u will end up with -25/(x-2)</p>