The average value of 𝑓(𝑥)=𝑥2𝑥3+1 on the closed interval [0,2] is

The average value of a function f(x) on the interval [a, b] is given by the formula:

1/(b-a) * ∫ from a to b f(x) dx

Here, our function f(x) = x^2/(x^3 + 1) and the interval is [0, 2]. So, we need to compute the integral of f(x) from 0 to 2 and then divide by 2 - 0 = 2.

The integral ∫ from 0 to 2 f(x) dx = ∫ from 0 to 2 x^2/(x^3 + 1) dx.

This integral is a bit tricky to compute directly, so we can use a substitution. Let u = x^3 + 1. Then du = 3x^2 dx, and dx = du/(3x^2).

Substituting these into the integral, we get:

∫ from 1 to 9 1/(3u) du = 1/3 * ∫ from 1 to 9 1/u du = 1/3 * [ln|u|] from 1 to 9 = 1/3 * (ln|9| - ln|1|) = 1/3 * ln(9).

So, the average value of f(x) on [0, 2] is:

1/2 * 1/3 * ln(9) = ln(9)/6.

To find the average value of a function on a closed interval [a, b], we use the formula:

Average value = (1/(b-a)) * ∫ from a to b f(x) dx

Here, f(x) = x^2/(x^3 + 1), a = 0, and b = 2.

So, we need to calculate the integral of f(x) from 0 to 2 and then divide it by 2-0 = 2.

The integral of f(x) = x^2/(x^3 + 1) from 0 to 2 can be calculated using a standard table of integrals or a computer algebra system.

Let's denote the integral as I.

I = ∫ from 0 to 2 x^2/(x^3 + 1) dx

To solve this integral, we can use the substitution method. Let u = x^3 + 1, then du = 3x^2 dx.

So, the integral becomes:

I = ∫ from 1 to 9 (1/3) du/u = (1/3) ln|u| from 1 to 9 = (1/3) ln|9| - (1/3) ln|1| = (1/3) ln(9)

So, the average value of the function on the interval [0, 2] is:

Average value = (1/2) * I = (1/2) * (1/3) ln(9) = (1/6) ln(9)