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What is the average value of a function #sin(X)# on the interval [0, pi]?

Answer 1

Emily Ammerman

#2/pi#.

Average# =(int sin x dx) /pi#, over the interval #[0, pi]#

The denominator is the length of the interval of integration.t.

#=[-cos x]/pi#, between the limits [0, pi]#

#--[cos pi - cos 0]#

#=-[-1-1]/pi=2/pi#.

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Answer 2

Christopher Allers

To find the average value of a function ( \sin(x) ) on the interval ([0, \pi]), we use the formula for the average value of a function over an interval:

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

Plugging in the values ( a = 0 ) and ( b = \pi ) for the interval and ( f(x) = \sin(x) ) for the function, we have:

[ \text{Average value} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin(x) , dx ]

[ = \frac{1}{\pi} \left( -\cos(x) \right) \bigg|_{0}^{\pi} ]

[ = \frac{1}{\pi} \left( -\cos(\pi) - (-\cos(0)) \right) ]

[ = \frac{1}{\pi} \left( 1 - (-1) \right) ]

[ = \frac{2}{\pi} ]

So, the average value of ( \sin(x) ) on the interval ([0, \pi]) is ( \frac{2}{\pi} ).

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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.