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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 $\displaystyle \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