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

Question

Scarlett Allison · Accepted Answer

#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#)

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

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

Graphical help

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#

Adam Adkins · Answer

undefined To find the net area between $ f(x) = x - \sin(x) $ and the x-axis over $ x $ in $[0, 3\pi]$, you need to compute the definite integral of $ |f(x)| $ over the given interval. This is because the graph of $ f(x) $ dips below the x-axis, and we're interested in the total area between the curve and the x-axis.

$$ \text{Net area} = \int_{0}^{3\pi} |f(x)| \, dx $$

$$ = \int_{0}^{3\pi} |x - \sin(x)| \, dx $$

You can evaluate this integral numerically using methods like numerical integration or software tools.