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.`
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