How do you find the average value of the function for #f(x)=sinxcosx, 0<=x<=pi/2#?

1 Answer
Feb 26, 2017

The average value is #1/pi#.

Explanation:

The average value of a function #f(x)# on a closed interval #[a, b]# is given by

#A = 1/(b - a)int_a^b F(x)#

Where #A# is the average value and #f'(x) = F(x)#. Our expression will therefore be

#A = 1/(pi/2 - 0) int_0^(pi/2) sinxcosx dx#

#A = 2/piint_0^(pi/2) sinxcosxdx#

The trick here is to realize that #2sinxcosx = sin2x#. Therefore, #sinxcosx = 1/2sin2x#.

#A = 2/piint_0^(pi/2) 1/2sin2xdx#

Now let #u = 2x#. Then #du = 2dx# and #dx = (du)/2#. The bounds of integration become #0# to #pi#.

#A = 2/piint_0^(pi) 1/4sinu du#

#A = 1/4(2/pi)int_0^(pi)sinudu#

#A = 1/(2pi)int_0^(pi)sinudu#

#A = 1/(2pi)[-cosu]_0^(pi)#

#A = 1/(2pi)[-cos2x]_0^(pi/2)#

#A = 1/(2pi)(-cospi - (-cos0))#

#A = 1/(2pi)(1 + 1)#

#A = 1/pi#

Hopefully this helps!