What is the average value of a function #y=7 sin x# on the interval #[0,pi]#?

Answer 1

The answer is #=14/pi#

The average value of a function #y=f(x)# from #x=a# to #x=b# is
#bary=(int_a^bf(x)dx)/(b-a)#

Therefore,

#bary=(int_0^(pi)7sinxdx)/(pi-0)#
#=7/pi[-cosx]_0^pi #
#=7/pi[1+1] #
#=14/pi#
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Answer 2

To find the average value of a function ( y = 7 \sin x ) on the interval ([0, \pi]), you can 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 ]

Where:

  • (a) and (b) are the endpoints of the interval.
  • (f(x)) is the function.

Substitute (a = 0), (b = \pi), and (f(x) = 7 \sin x) into the formula:

[ \text{Average value} = \frac{1}{\pi - 0} \int_0^{\pi} 7 \sin x , dx ]

Then, integrate (7 \sin x) from (0) to (\pi):

[ \int_0^{\pi} 7 \sin x , dx = -7 \cos x \Big|_0^{\pi} = -7 \cos(\pi) + 7 \cos(0) = -7(-1) + 7(1) = 14 ]

Finally, divide the result by (\pi - 0 = \pi):

[ \text{Average value} = \frac{1}{\pi} \times 14 = \frac{14}{\pi} ]

So, the average value of (y = 7 \sin x) on the interval ([0, \pi]) is (\frac{14}{\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.

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.

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.

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.

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