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

##### 1 Answer

#### Explanation:

The average value

#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

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

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

Splitting up

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

This is a common integral (note that

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

Evaluating:

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

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

Note that

#c=1/2sin(2)#

#c approx0.4546487#