VCAA multiple choice q22

[croc76]

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

[chopz]

inverse both equations, much easier
Then solve normally once you've inversed them

[jto1292]

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..

[croc76]

hmmm... what would be the way to do it without using an inverse?

[jasoN-]

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.




switch x and y



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

[croc76]

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??

[croc76]

actually my bad, that doesnt work. what would the graph of the area look like?

im just having a hard time visualizing this

thanks

[Whatlol]

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.

[croc76]

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