What is the average value of the function #f(x)=cos(x/2) # on the interval #[-4,0]#?

1 Answer
Aug 29, 2017

#1/2sin(2)#, approximately #0.4546487#

Explanation:

The average value #c# of a function #f# on the interval #[a,b]# is given by:

#c=1/(b-a)int_a^bf(x)dx#

Here, this translates into the average value of:

#c=1/(0-(-4))int_(-4)^0cos(x/2)dx#

Let's use the substitution #u=x/2#. This implies that #du=1/2dx#. We can then rewrite the integral as such:

#c=1/4int_(-4)^0cos(x/2)dx#

#c=1/2int_(-4)^0cos(x/2)(1/2dx)#

Splitting up #1/4# into #1/2*1/2# allows for #1/2dx# to be present in the integral so we can easily make the substitution #1/2dx=du#. We also need to change the bounds into bounds of #u#, not #x#. To do this, take the current #x# bounds and plug them into #u=x/2#.

#c=1/2int_(-2)^0cos(u)du#

This is a common integral (note that #d/dxsin(x)=cos(x)#):

#c=1/2[sin(u)]_(-2)^0#

Evaluating:

#c=1/2(sin(0)-sin(-2))#

#c=-1/2sin(-2)#

Note that #sin(-x)=-sin(x)#:

#c=1/2sin(2)#

#c approx0.4546487#