How do you find the area under one half period of #y=2sin3x#?

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Answer 1 · 2017-03-23T20:48:30

#A=4/3#

We can set the lower limit of integration at #x=0#. The half period is then the interval to:

#3x = pi#

that is:

#x=pi/3#

Then:

#A = int_0^(pi/3) 2 sin3x dx = 2/3 int_0^(pi/3) sin3x d(3x)#

#A=2/3 [-cos3x]_0^(pi/3) = 4/3#