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

1 Answer
mason m
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

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