To find the mean (average) value of a function $f(x)$ over an interval $[a,b]$, we use the formula:

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

In this case, the function is $f(x) = \frac{3}{4}(x-1)(3-x)$ and the interval is $[1,3]$. Let's calculate the mean step by step:

Step 1: Set up the formula

The interval is $[1, 3]$, so we substitute $a = 1$ and $b = 3$:

$\text{Mean value of } f(x) = \frac{1}{3-1} \int_{1}^{3} \frac{3}{4}(x-1)(3-x) \, dx$

This simplifies to:

$\text{Mean value of } f(x) = \frac{1}{2} \int_{1}^{3} \frac{3}{4}(x-1)(3-x) \, dx$

Step 2: Expand the function

Now let's expand $(x-1)(3-x)$:

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

Thus, the function becomes:

$f(x) = \frac{3}{4}(-x^2 + 4x - 3)$

Step 3: Integrate the function

Now, we integrate $f(x)$ over the interval $[1,3]$:

$\int_{1}^{3} \frac{3}{4}(-x^2 + 4x - 3) \, dx = \frac{3}{4} \int_{1}^{3} (-x^2 + 4x - 3) \, dx$

We integrate each term separately:

$\int_{1}^{3} -x^2 \, dx = \left[ -\frac{x^3}{3} \right]_{1}^{3} = -\frac{27}{3} + \frac{1}{3} = -9 + \frac{1}{3} = -\frac{26}{3}$

$\int_{1}^{3} 4x \, dx = \left[ 2x^2 \right]_{1}^{3} = 2(9) - 2(1) = 18 - 2 = 16$

$\int_{1}^{3} -3 \, dx = \left[ -3x \right]_{1}^{3} = -3(3) + 3(1) = -9 + 3 = -6$

Now, putting these together:

$\frac{3}{4} \left( -\frac{26}{3} + 16 - 6 \right) = \frac{3}{4} \left( -\frac{26}{3} + 10 \right)$

Convert 10 to a fraction:

$\frac{3}{4} \left( -\frac{26}{3} + \frac{30}{3} \right) = \frac{3}{4} \left( \frac{4}{3} \right)$

Simplifying:

$\frac{3}{4} \times \frac{4}{3} = 1$

Step 4: Calculate the mean value

Since the integral evaluates to 1, the mean value is:

$\text{Mean value of } f(x) = \frac{1}{2} \times 1 = \frac{1}{2}$

Thus, the mean value of $f(x)$ over the interval $[1, 3]$ is:

$\boxed{\frac{1}{2}}$

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

1. How does the mean value theorem apply to the function $f(x)$?
2. What is the significance of finding the mean value of a function?
3. How does changing the interval affect the mean of a function?
4. Can you find the mean of $f(x)$ over the interval $[0,2]$?
5. How would the result change if the function were $f(x) = \frac{1}{4}(x-1)(3-x)$?

Tip: When integrating polynomial functions, expanding them first can simplify the calculation.