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What is the average value of a function #y=secx tanx# on the interval #[0,pi/3]#?

Answer 1

William Abdalla

#3/pi#

The average value of the function #f(x)# on the interval #[a,b]# is equal to

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

Thus, here, the average value is

#1/(pi/3-0)int_0^(pi/3)secxtanxdx#

Note that #d/dx(secx)=secxtanx#, so #intsecxtanxdx=secx+C#.

#=1/(pi/3)[secx]_0^(pi/3)=3/pi(sec(pi/3)-sec(0))=3/pi(2-1)=3/pi#

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Answer 2

Luke Alvarez

To find the average value of the function ( y = \sec(x) \tan(x) ) on the interval ([0, \frac{\pi}{3}]), we use the formula for average value of a function over an interval:

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

Given ( a = 0 ) and ( b = \frac{\pi}{3} ), we can compute the integral:

[ \text{Average value} = \frac{1}{\frac{\pi}{3} - 0} \int_{0}^{\frac{\pi}{3}} \sec(x) \tan(x) , dx ]

After evaluating this integral, we obtain the average value of the function over the given interval.

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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.