What is the net area between #f(x) = x-sinx # and the x-axis over #x in [0, 3pi ]#?

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Answer 1 · 2018-05-07T22:11:32

#int_0^(3π)(x-sinx)dx=((9π^2)/2-2) m^2#

#f(x)=x-sinx# , #x##in##[0,3pi]#

#f(x)=0# #<=># #x=sinx# #<=># #(x=0)#

(Note: #|sinx|<=|x|# , #AA##x##in##RR# and the #=# is true only for #x=0#)

So when #x##in##[0,3pi]# , #f(x)>=0#

Graphical help enter image source here

The area we are looking for since #f(x)>=0# ,#x##in##[0,3pi]#

is given by #int_0^(3π)(x-sinx)dx# #=#

#int_0^(3π)xdx# #- int_0^(3π)sinxdx# #=#

#[x^2/2]_0^(3π)+[cosx]_0^(3π)# #=#

#(9π^2)/2+cos(3π)-cos0# #=#

#((9π^2)/2-2)# #m^2#