Wednesday, October 6, 2010

`y = 1/x, y = x, y = (1/4)x, x>0` Sketch the region enclosed by the given curves and find its area.

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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