What is the net area between #f(x)=2/x-x# in #x in[1,2] # and the x-axis?

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Answer 1 · 2017-04-07T17:25:26

#A = 1/2#

There is a zero crossing within the interval:

#0 = 2/x-x#

#x^2 = 2#

#x = sqrt(2)#

The area will be positive #1 <=x<=sqrt(2)# and negative #sqrt(2) < x <=2# Therefore, we subtract the integral in the second region from the first:

A = #int_1^sqrt(2)(2/x-x)dx-int_sqrt(2)^2(2/x-x)dx=#

#(2ln|x|-x^2/2]_1^sqrt2- (2ln|x|-x^2/2]_sqrt2^2=#

#(ln(x^2)-x^2/2]_1^sqrt2- (ln(x^2)-x^2/2]_sqrt2^2=#

#(ln(2)-2/2- ln(1)+1/2)- (ln(4)-4/2- ln(2)+2/2)=#

#ln(2)-1- 0+1/2-ln(4)+2+ln(2)-1= 1/2#