Find the mean of f(x) = 3/4(x-1)(3-x) for x = [1,3]

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.

Explore how to find the mean value of the function f(x) = 3/4(x-1)(3-x) over the interval [1,3]. This step-by-step guide covers integral calculus, polynomial expansion, and the application of the mean value theorem. Ideal for undergraduate students.

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