This is an area question of the type:
#A=int_a^b f(x) dx - int_a^b g(x) dx#
Where #f(x) = 2# and #g(x) = 2sin(pix)# and #a=1# and #b=2#.
First of all it helps to find out the points of intersection for #f# and #g#.
Letting #2=2sin(pix)#, we find that #x=1/2#, which is irrelevant as the integral is between #1# and #2#.
Next we antidifferentiate #f(x)#:
#int 2 dx = 2x#
Antidifferentiate #g(x)#:
#int 2sin(pix) dx = (-2cos(pix))/pi#
The coefficient of #-2# can be disregarded. By the chain rule we can see that the the #pi# in the denominator is eliminated when multiplying by the #pi# in the argument of the function, i.e.
#d/dx(-2cos(pix)) = d/dx(pix)*d/dx(-2cos(u))# where #u=pix#
This derivative becomes #2pisin(pix)#, hence the division by #pi# is necessary.
Finally to integrate:
The area between #f(x)# and #g(x)# is given by:
#int_1^2 2 dx - int_1^2 2sin(pix) dx#
#=[2x]_1^2 - [(-2cos(pix))/pi]_1^2#
#=4-2 - (-2/pi - 2/pi)#
#=2 + 4/pi#
I hope this helps!
