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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: croc76 on November 07, 2010, 10:48:55 am

Title: VCAA multiple choice q22
Post by: croc76 on November 07, 2010, 10:48:55 am: So i really really dont understand this one. anyone please explain it?

consider the region bounded by the x-axis, the y-axis , the line with equation y=3 and the curve with the equation ln(x-1)

how is this done?
thanks
Title: Re: VCAA multiple choice q22
Post by: chopz on November 07, 2010, 11:02:20 am: inverse both equations, much easier :)
Then solve normally once you've inversed them
Title: Re: VCAA multiple choice q22
Post by: jto1292 on November 07, 2010, 11:06:31 am: Do an integration with y=3 -> y=0 and the equation x= e^y + 1.. Much easier then trying to find x-axis bound region and then subtracting..
Title: Re: VCAA multiple choice q22
Post by: croc76 on November 07, 2010, 11:24:35 am: hmmm... what would be the way to do it without using an inverse?
Title: Re: VCAA multiple choice q22
Post by: jasoN- on November 07, 2010, 11:26:41 am: A quick sketch won't hurt.
When you sketch the diagram, you will see that you need to find the area between the line y=3 and the graph ln(x-1).
Obviously you cannot antidifferentiate log graphs (for now), so an alternative is to find the inverse of the graph.

$y = log_e(x-1)$
$e^y = x-1$
$e^y + 1 = x$
switch x and y
$y^{-1} = e^x + 1$
$\int_0^{y=3} log_e (x-1) dx = \int_0^{x=3} e^x + 1 dx$

I don't know if there are any other ways, but these types of questions come up quite a bit, so it's good to know the inverse way
Title: Re: VCAA multiple choice q22
Post by: croc76 on November 07, 2010, 11:30:59 am: Is it possible to find the coordinate where 3 =ln(x-1), find the area under y=3 till x=2 then integrate (3-ln(x-1)) for x=2 to the point where y=3 intercepts ln (x-1)

that would work right??
Title: Re: VCAA multiple choice q22
Post by: croc76 on November 07, 2010, 11:33:33 am: actually my bad, that doesnt work. what would the graph of the area look like?

im just having a hard time visualizing this

thanks
Title: Re: VCAA multiple choice q22
Post by: Whatlol on November 07, 2010, 11:36:49 am: Quote from: croc76 on November 07, 2010, 11:30:59 am
Is it possible to find the coordinate where 3 =ln(x-1), find the area under y=3 till x=2 then integrate (3-ln(x-1)) for x=2 to the point where y=3 intercepts ln (x-1)

that would work right??

yes you can find the area under the graph from x =2 to x = e^3 +1.

but then you must do 3(e^3 +1) - that area.
Title: Re: VCAA multiple choice q22
Post by: croc76 on November 07, 2010, 11:37:09 am: Wait its all good i got it!!!

so 6 +(e^3 -4)

6 is the area under y=3 until ln intercepts the x axis. and e^3-4 is antidiff(3-ln(x-1),x,2,e^(3)+1)

thanks guys