Average of sin t *cos t how to find ?
1 Answer
Apr 29, 2018
# bar (f)(t) = 1/pi ~~ 0.318 #
Explanation:
By definition, the average value of a function
# bar (f)(x) = 1/(b-a) \ int_a^b \ f(x) \ dx #
So, Consider the function:
# f(t)=sintcost #
Which, using the identity
# f(t)=1/2sin(2t)#
An interval is not specified (and if we considered an entire period, them by symmetry, we would get
# sin(2t) = 0 => 2t=0,pi/2 => t=0, pi/4#
Thus for a single positive cycle, we have:
# bar (f)(t) = 1/(pi/4-0) \ int_(0)^(pi/4) \ 1/2sin(2t) \ dt #
# \ \ \ \ \ \ \ = (1/2)/(pi/4) \ [-1/2 cos(2t)]_(0)^(pi/4) #
# \ \ \ \ \ \ \ = -1/pi \ (0-1) #
# \ \ \ \ \ \ \ = 1/pi #
# \ \ \ \ \ \ \ ~~ 0.318 #