
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!

1 Attachment(s)
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: