Question

What is the average value of the function `f(x) = 4x ln (2x)` on the interval `[1,4]`?

Answer 1

`94/3ln(2)-5`

Explanation:

The average value `s` of a function `f` on the interval `[a,b]` can be found through the formula

`s=1/(b-a)int_a^bf(x)dx`

So here, the average value is equal to

`s=1/(4-1)int_1^4 4xln(2x)dx`

Let's first find this integral without the bounds.

`I=int4xln(2x)dx`

To do this integral, we'll need to use integration by parts. This technique takes the form `intudv=uv-intvdu`. Here, we want to choose a value of `u` that becomes simpler when differentiated and a value of `dv` that can be integrated.

For the given integral, let `u=ln(2x)` and `dv=4xcolor(white).dx`. Differentiating `u` and integrating `dv` we see that:

`{(u=ln(2x),=>,(du)/dx=1/x,=>,du=1/xdx),(dv=4xcolor(white).dx,=>,intdv=int4xcolor(white).dx,=>,v=2x^2):}`

So:

`I=uv-intvdu=2x^2ln(2x)-int2x^2 1/xdx`

Simplifying and integrating:

`I=2x^2ln(2x)-int2xcolor(white).dx=2x^2ln(2x)-x^2=x^2(2ln(2x)-1)`

This is without the constant go integration because we will use this to evaluate the integral. Returning to the average value, we see that

`s=1/(4-1)int_1^4 4xln(2x)dx=1/3[x^2(2ln(2x)-1)]_1^4`

Now evaluating in full:

`s=1/3 16(2ln(8)-1)-1/3 1(2ln(2)-1)`

`s=32/3ln(8)-16/3-2/3ln(2)+1/3`

Note that `ln(8)=ln(2^3)=3ln(2)`:

`s=32ln(2)-2/3ln(2)-16/3+1/3`

`s=94/3ln(2)-5`