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

Answer 1

Christopher Allers

#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#

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Answer 2

Adam Adkins

To find the area under one half period of ( y = 2\sin(3x) ):

Identify one half period of the function ( y = 2\sin(3x) ). The period of ( \sin(3x) ) is ( \frac{2\pi}{3} ) because the coefficient of ( x ) inside the sine function affects the period. Therefore, half of the period is ( \frac{\pi}{3} ).
Set up the definite integral to calculate the area under the curve over one half period: [ \text{Area} = \int_{0}^{\pi/3} 2\sin(3x) , dx ]
Integrate the function ( 2\sin(3x) ) with respect to ( x ) over the interval ([0, \pi/3]) to find the area.
Evaluate the integral using the antiderivative of ( 2\sin(3x) ).

By calculating this definite integral, you will find the area under one half period of the curve ( y = 2\sin(3x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.