What is the average of (x)/(5x^2+3)^2 on the interval [0,2]?

Question

I know that the average of f(x)= 1/(b-a) x integral from a to b of f(x) but i cant figure out the correct way to integrate.

Answer 1

Let f(x) = x/(5x^2 + 3)^2

The average is evaluated by the following formula:

A = [1/(b - a)] Integral (a to b, f(x) dx)

Since we want the average on the interval 0 to 2, a = 0 and
b = 2, and our integral is

A = [1/(2 - 0)] Integral (0 to 2, x/(5x^2 + 3)^2 dx)

I'm going to rearrange this, to make the next step obvious.

A = [1/2] Integral (0 to 2, (1/(5x^2 + 3)^2) x dx)

To solve this integral, use substitution. Note that we can simultaneously change our bounds for integration upon using substitution.

Let u = 5x^2 + 3. To solve for the new bounds, plug in our current bounds to find out what function we get.
When x = 0, u = 5(0)^3 + 3 = 3
When x = 2, u = 5(2)^3 + 3 = 5(8) + 3 = 43.
So our new bounds is 3 to 43.

du = 10x dx, so
(1/10)du = x dx.

x dx is the tail end of our current integral, so it subsequently follows that (1/10)du is going to be the tail end of our new integral.

A = [1/2] Integral (3 to 43, (1/u^2)(1/10) du )

Pull out the constant (1/10), and it will merge with (1/2) to become (1/20).

A = (1/20) Integral (3 to 43, (1/u^2) du )

The integral of 1/u^2 is -1/u (verifiable through the reverse power rule), so we evaluate this integral.

A = (1/20) [-1/u] {evaluated from 3 to 43}

A = (1/20) [-1/43 - (-1/3)]

A = (1/20) [-1/43 + 4/3]

A = (1/20) [-3/129 + 172/129]

A = (1/20) [169/129]

A = 169/2580