Thread: Areas Between Curves

Find the area enclosed by the 3 lines:

y = 1/x, y = x, and y = x/4 where x > 0

The answer is supposed to be LN 2

I know that one of the first steps is to solve the equations simultaneously, but I'm not sure how to so what when you're working with 3 equations. Anyway, this is how I did it.

1/x = x

1/x -x = x/4

1/x - x - x/4 = 0

x = +/- 2/sqrt(5)

So I integratd between +/- 2/sqrt(5) and I did not get LN 2.

Can someone help me with this, please?

You have 2 lines, y=x, and y=x/4, and a curve y=1/x.
You can not fin only one intersection point.
You have to find the intersection point using 2 equations at a time.
The two line, a line and the curve, the other line and the curve.
Try to plot the graph to realy see your area.
After that try to integrate.
Hope this helps!

ok
here you go,
y=x and y=1/x
x=1/x, so x^2=1, since x>0, x=1 is the intersection point.

y=x and y=x/4
x=x/4
4x=x
3x=0
x=0

y=1/x, y=x/4
1/x=x/4
x^2=4, since x>0, x=2

Graph the functions: