How do you find the area under one half period of #y=2sin3x#?
that is:
Then:
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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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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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 Calculus 2. Is this the right answer?
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