Question

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

Answer 1

`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`