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What is the average value of the function #f(x)=sin4x# on the interval #[-pi,pi]#?

Answer 1

Isaiah Allen

It is #0#.

The average value of an continuous function #f# on interval #[a,b]# is

#1/(b-a) int_a^b f(x) dx#.

Here, we have #1/(pi-(-pi)) int_(-pi)^pi sin4x dx =1/(2pi) [-1/4 cos(4x)]_-pi^pi#

# = - 1/(8pi) [cos(4pi)-cos(-4pi)] = 1/(8pi) [0] = 0#

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Answer 2

Elijah Abram

To find the average value of the function (f(x) = \sin(4x)) on the interval ([- \pi, \pi]), we use the formula:

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

In this case, (a = -\pi) and (b = \pi). So, we have:

[ \text{Average value} = \frac{1}{\pi - (-\pi)} \int_{-\pi}^{\pi} \sin(4x) , dx ]

[ = \frac{1}{2\pi} \left[ -\frac{\cos(4x)}{4} \right]_{-\pi}^{\pi} ]

[ = \frac{1}{2\pi} \left[ -\frac{\cos(4\pi)}{4} + \frac{\cos(-4\pi)}{4} \right] ]

[ = \frac{1}{2\pi} \left[ -\frac{1}{4} + \frac{1}{4} \right] ]

[ = \frac{1}{2\pi} \times 0 ]

[ = 0 ]

Therefore, the average value of the function (f(x) = \sin(4x)) on the interval ([- \pi, \pi]) is (0).

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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.