To find the mean of the function $f(x) = 4x(1 - x^2)$ over the interval $[0, 1]$, we use the formula for the mean value of a function on a closed interval:

$\text{Mean} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx$

where $a = 0$ and $b = 1$. Thus, the mean becomes:

$\text{Mean} = \int_{0}^{1} 4x(1 - x^2) \, dx$

Step 1: Expand the integrand

First, expand $4x(1 - x^2)$:

$4x(1 - x^2) = 4x - 4x^3$

Step 2: Integrate term-by-term

Now, integrate each term separately:

$\int_{0}^{1} (4x - 4x^3) \, dx = \int_{0}^{1} 4x \, dx - \int_{0}^{1} 4x^3 \, dx$

Calculate these integrals:

1. $\int_{0}^{1} 4x \, dx = 4 \left[ \frac{x^2}{2} \right]_{0}^{1} = 4 \cdot \frac{1}{2} = 2$

2. $\int_{0}^{1} 4x^3 \, dx = 4 \left[ \frac{x^4}{4} \right]_{0}^{1} = 4 \cdot \frac{1}{4} = 1$

Step 3: Subtract the results

The integral from 0 to 1 of $4x(1 - x^2)$ is:

$2 - 1 = 1$

Step 4: Divide by the length of the interval

The length of the interval $[0,1]$ is $1$, so the mean is:

$\text{Mean} = \frac{1}{1} \cdot 1 = 1$

Therefore, the mean value of the function $f(x) = 4x(1 - x^2)$ over the interval $[0,1]$ is $1$.

Would you like more details or have any questions?

Here are some related questions:

1. How would you find the mean for a different interval, such as $[0,2]$?
2. What would the mean be if the function was $f(x) = 4x^2(1 - x^2)$?
3. Can you find the average rate of change of $f(x)$ over $[0,1]$?
4. What is the maximum value of $f(x)$ in the interval $[0,1]$?
5. How would you find the median value of $f(x)$ over the interval?

Tip: When integrating polynomial functions, break them down into simpler terms and integrate term-by-term for ease.