#(cc (a))# Saying that the graph of a function #f# lays above or on the x-axis over some interval #[alpha,beta]# means proving that
#f(x) >= 0#, #forall x in [alpha,beta]#
Hence we have to solve the inequality
#-(x+1)sin(5x)>=0#
As #x in [(7pi)/5, (8pi)/5]#, #x+1# is always going to be positive, meaning that #-(x+1)# is negative. In order for their product to be positive, #sin(5x)# also has to be negative.
#:. sin(5x) <= 0#
#x in [(7pi)/5,(8pi)/5]=> 5x in [7pi,8pi]#
This interval starts with an odd multiple of #pi# and ends with an even one, hence it represents the lower part of the unit circle, where sine is negative.
#=> sin(5x) <= 0=> -(x+1)sin(5x)>=0#
#=> f(x) >=0 forall x in [(7pi)/5,(8pi)/5]#
#(cc b)# The area between the graph of a function #f# and the x-axis between some bounds #x_1# and #x_2# is given by the definite integral
#"Area"=int_(x_1)^(x_2) f(x)"d"x#
:. #"Area" = int_(7pi"/"5)^(8pi"/"5) -(x+1)sin(5x) "d"x#
#(cc c)# Integration by parts states that, for some functions #g# and #h#,
#int g(x)h'(x)"d"x = [g(x)h(x)]-int g'(x)h(x)"d"x#
We will be able to find the definite integral using the antiderivative.
Let #g(x) = x+1# and #h'(x) = -sin(5x)#.
#g(x) = x+1 => g'(x) = 1#
#h'(x)=-sin(5x)=>h(x)=int -sin(5x)"d"x#
Substitute #u=5x#, #"d"x=("d"u)/5# to get the standard integral of #sinu#:
#h(x) = cos(5x)/5 + C#
In integration by parts, we take #C# to be #0#.
Let #I=inth(x) "d"x=intcos(5x)/5"d"x=1/5intcos(5x)"d"x#.
Let #u=5x => "d"x=("d"u)/5#.
#I=1/5intcosu("d"u)/5=1/25 intcos(u)"d"u=sinu/25=sin(5x)/25#
Again, #C=0#.
#int-(x+1)sin(5x)"d"x=((x+1)cos(5x))/5-sin(5x)/25#
Therefore,
#int_(7pi"/"5)^(8pi"/"5) -(x+1)sin(5x)"d"x#
# = [((x+1)cos(5x))/5-sin(5x)/25]_(7pi"/"5)^(8pi"/"5)#
# = (((8pi)/5+1)cos(8pi))/5-sin(8pi)/25-(((7pi)/5+1)cos(7pi))/5+sin(7pi)/25#
# = (8pi+5)/25+(7pi+5)/25=(15pi+10)/25=(3pi+2)/5#
