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

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Answer 1 · 2016-10-08T12:50:34

#(16-16sqrt2)/pi#

The average value of the function #f(t)# on the interval #[a,b]# is:

#1/(b-a)int_a^bf(t)dt#

Thus, where #f(t)=-4sec(t)tan(t)# and the interval is #[0,pi/4]#, we see its average value on that interval is:

#1/(pi/4-0)int_0^(pi/4)-4sec(t)tan(t)dt#

#=(-4)/(pi/4)int_0^(pi/4)sec(t)tan(t)dt#

Note that since #d/dtsec(t)=sec(t)tan(t)#, so the integral of #sec(t)tan(t)# is #sec(t)+C#.

#=-16/pi[sec(t)]_0^(pi/4)#

#=-16/pi[sec(pi/4)-sec(0)]#

Note that #cos(pi/4)=sqrt2/2#, so #sec(pi/4)=2/sqrt2=sqrt2#.

#=-16/pi(sqrt2-1)#

#=(16(1-sqrt2))/pi#