# Finding Area Enclosed by Three Curves

• January 24th 2013, 08:39 AM
shennara123
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!
• January 24th 2013, 09:10 AM
Plato
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 $\int_0^{2/3} {\left[ {6x - 3x} \right]dx} + \int_{2/3}^1 {\left[ {\frac{2}{x} + 1 - 3x} \right]dx}$

Look at the graph