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 #f(x)# over a domain #[a,b]# is given by:

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

So, Consider the function:

# f(t)=sintcost #

Which, using the identity #sin2A -= 2sinAcosA#, we can write

# f(t)=1/2sin(2t)#

An interval is not specified (and if we considered an entire period, them by symmetry, we would get #0#, so we consider a single positive cycle.

# 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 #