What is the average value of a function #sec^2x# on the interval #[pi/6, pi/4]#?
1 Answer
Jun 6, 2016
Explanation:
The average value of the function
#1/(b-a)int_a^bf(x)dx#
Thus, the average value of
#1/(pi/4-pi/6)int_(pi/6)^(pi/4)sec^2xdx#
Note that
Also, note that
The average value then equals:
#1/(pi/4-pi/6)int_(pi/6)^(pi/4)sec^2xdx=12/pi[tanx]_(pi/6)^(pi/4)#
#=12/pi[tan(pi/4)-tan(pi/6)]=12/pi[1-sqrt3/3]=12/pi[(3-sqrt3)/3]#
#=4/pi[3-sqrt3]=(12-4sqrt3)/pi#