What is the average value of the function #f(x) = sec x tan x# on the interval #[0,pi/4]#?

Answer 1

Isabella Ackerman

It is # (4(sqrt2-1))/pi#

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

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

So the value we seek is

#1/(pi/4-0) int_0^(pi/4) secxtanx dx#

# = 4/pi [secx]_0^(pi/4)#

# = 4/pi [sec(pi/4)-sec(0)]#

# = 4/pi [sqrt2-1]#

# = (4(sqrt2-1))/pi#

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Answer 2

Evelyn Agard

To find the average value of the function ( f(x) = \sec(x) \tan(x) ) on the interval ([0, \frac{\pi}{4}]), follow these steps:

Calculate the definite integral of the function ( f(x) ) over the given interval.
Divide the result from step 1 by the length of the interval ([0, \frac{\pi}{4}]).
The result obtained in step 2 is the average value of the function over the interval.

This process gives the average value of the function ( f(x) = \sec(x) \tan(x) ) on the interval ([0, \frac{\pi}{4}]).

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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.