We know that the lower and upper bounds are #0, 3pi# and that the function is #f(x)=sin(x)-cos(x)#, so we can rewrite the question as the definite integral of #sin(x)-cos(x)# from #0# to #3pi#.
#int_0^(3pi)sin(x)-cos(x)dx#
We know that the anti-derivative of #sin(x)# is #-cos(x)# and that the anti-derivative of #cos(x)# is equal to #sin(x)#. So we can find the anti-derivative of #sin(x) - cos(x)# by substituting their anti-derivatives to get that #F(x) = -cos(x) - sin(x)#.
We can then evaluate the anti-derivative at its bounds #0, 3pi#.
So the area #A# is equal to #F(3pi) - F(0)#
#F(3pi) = -cos(3pi) - sin(3pi) = 1#
#F(0) = -cos(0) - sin(0) = -1#
So the area #A# is equal to #1 - (-1) = 2#
So #int_0^(3pi)sin(x)-cos(x)dx = 2#.
