Find the average value of the function f(x) = |x| - 1 over [-1, 3].

Steps to Solve

1. Determine the average value formula

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

$$ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) , dx $$

For this problem, we have $a = -1$ and $b = 3$.

2. Calculate the width of the interval

First, we find the width of the interval:

$$ b - a = 3 - (-1) = 4 $$

3. Set up the integral for the function

Next, we need to integrate the function $f(x) = |x| - 1$. The function has two cases in the interval $[-1, 3]$:

• For $x \in [-1, 0]$: $f(x) = -x - 1$
• For $x \in [0, 3]$: $f(x) = x - 1$

Now, break the integral into the two parts:

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

4. Calculate the first integral

Calculate the first part:

$$ \int_{-1}^0 (-x - 1) , dx = \left[ -\frac{x^2}{2} - x \right]_{-1}^0 $$

Evaluating this, we get:

$$ \left[ 0 - 0 \right] - \left[ -\frac{1}{2} + 1 \right] = 0 + \frac{1}{2} = \frac{1}{2} $$

5. Calculate the second integral

Calculate the second part:

$$ \int_{0}^3 (x - 1) , dx = \left[ \frac{x^2}{2} - x \right]_{0}^{3} $$

Evaluating this, we get:

$$ \left[ \frac{9}{2} - 3 \right] - \left[ 0 - 0 \right] = \frac{9}{2} - 3 = \frac{3}{2} $$

6. Combine the integrals

Now combine the results of both integrals:

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

7. Calculate the average value

Finally, plug this value into the average value formula:

$$ \text{Average value} = \frac{1}{4} \cdot 2 = \frac{1}{2} $$