What is the average value of a function #f(t)=−4sec(t)tan(t)# on the interval #[0, pi/4]#?
1 Answer
Oct 8, 2016
Explanation:
The average value of the function
#1/(b-a)int_a^bf(t)dt#
Thus, where
#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
#=-16/pi[sec(t)]_0^(pi/4)#
#=-16/pi[sec(pi/4)-sec(0)]#
Note that
#=-16/pi(sqrt2-1)#
#=(16(1-sqrt2))/pi#