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#