Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 27

Given y=1/x, y=x, y=1/4x, x>0
Find the intersection point of y=x and y=1/x.
x=1/x
x^2=1
x=1
When x=1, y=1. The intersection point is (1, 1).
Find the intersection point of y=1/4x and y=1/x.
1/4x=1/x
x^2=4
x=2
When x=2, y=1/2. The intersection point is (2, 1/2).

A=int_0^1(x-1/4x)dx+int_1^2(1/x+1/4x)dx
= int_0^1(3/4x)dx+int_1^2(1/x+1/4x)dx
=[3/4*x^2/2]_0^1+[lnx+1/4*x^2/2]_1^2
=[3/8x^2]_0^1+[lnx+1/8x^2]_1^2
=[3/8(1)^2-0]+[(ln2+1/8(2)^2)-(ln1+1/8(1)^2]
=3/8+ln2+1/2-ln1-1/8
=1/4+1/2+ln2-ln1
=3/4+ln2-ln1
=1.443
The area enclosed by the given curves is 1.443 units squared.
The curve in black is y=1/x.
The red line is y=x.
The green line is y=1/4x.