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Finding Area Enclosed by Three Curves
Find the area enclosed by
y=3x
y=6x
y=(2/x)+1
In the first quadrant of the Cartesian plane.
I have already point the points on intersection to be (0,0) (2/3,4) and (1,3)
I have posted a picture of my attempted solution, if anyone could tell me where I'm going wrong I'd really appreciate it.
Thanks!
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Re: Finding Area Enclosed by Three Curves
Quote:
Originally Posted by
shennara123 
Find the area enclosed by
y=3x
y=6x
y=(2/x)+1
In the first quadrant of the Cartesian plane.
Did you draw a graph?
It should be ![\int_0^{2/3} {\left[ {6x - 3x} \right]dx} + \int_{2/3}^1 {\left[ {\frac{2}{x} + 1 - 3x} \right]dx}](http://latex.codecogs.com/png.latex?\int_0^{2/3} {\left[ {6x - 3x} \right]dx} + \int_{2/3}^1 {\left[ {\frac{2}{x} + 1 - 3x} \right]dx} )
Look at the graph
