Pretty Good Prep

Understanding the Average Function Value

You already know how to find the average of a list of numbers: add them up and divide by how many there are. But what if you have a continuous function over an interval? You can't just "add up" infinitely many values!

The average value of a function extends the familiar averaging concept to continuous functions using integration. It answers: "If this function were constant, what single value would give the same total accumulation?"

The Formula

The average value of $f(x)$ on the interval $[a, b]$ is:

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

Why This Formula Makes Sense

Discrete average: $\text{average} = \frac{\text{sum of values}}{\text{number of values}}$

Continuous analog: $f_{\text{avg}} = \frac{\int_a^b f(x) \, dx}{b - a}$

The integral $\int_a^b f(x) \, dx$ is the "continuous sum" of all function values
The interval length $b - a$ is the "continuous count"

Geometric interpretation: The average value is the height of a rectangle with base $[a, b]$ that has the same area as the region under the curve.

Example 1: Average of a Linear Function

Problem: Find the average value of $f(x) = 2x + 1$ on $[0, 4]$.

$$f_{\text{avg}} = \frac{1}{4-0} \int_0^4 (2x + 1) \, dx$$

$$= \frac{1}{4} \left[ x^2 + x \right]_0^4$$

$$= \frac{1}{4} \left[ (16 + 4) - (0) \right]$$

$$= \frac{1}{4} \cdot 20 = 5$$

$$\boxed{f_{\text{avg}} = 5}$$

Check: For a linear function, the average equals the midpoint value. At $x = 2$: $f(2) = 2(2) + 1 = 5$ ✓

Example 2: Average of a Quadratic

Problem: Find the average value of $f(x) = x^2$ on $[0, 3]$.

$$f_{\text{avg}} = \frac{1}{3-0} \int_0^3 x^2 \, dx$$

$$= \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3$$

$$= \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3$$

$$\boxed{f_{\text{avg}} = 3}$$

Note: The midpoint value is $f(1.5) = 2.25$, but the average is $3$ because the function grows faster on the right side of the interval.

Example 3: Average of a Trig Function

Problem: Find the average value of $f(x) = \sin x$ on $[0, \pi]$.

$$f_{\text{avg}} = \frac{1}{\pi - 0} \int_0^{\pi} \sin x \, dx$$

$$= \frac{1}{\pi} \left[ -\cos x \right]_0^{\pi}$$

$$= \frac{1}{\pi} \left[ -\cos(\pi) - (-\cos(0)) \right]$$

$$= \frac{1}{\pi} \left[ -(-1) + 1 \right]$$

$$= \frac{1}{\pi} \cdot 2 = \frac{2}{\pi}$$

$$\boxed{f_{\text{avg}} = \frac{2}{\pi} \approx 0.637}$$

Example 4: Average Temperature

Problem: The temperature (in °F) over a 12-hour period is modeled by $T(t) = 60 + 10\sin\left(\frac{\pi t}{12}\right)$ where $t$ is hours after 6 AM. Find the average temperature from 6 AM to 6 PM.

We need the average of $T(t)$ on $[0, 12]$.

$$T_{\text{avg}} = \frac{1}{12-0} \int_0^{12} \left(60 + 10\sin\frac{\pi t}{12}\right) dt$$

Split the integral:

$$= \frac{1}{12} \left[ 60t - 10 \cdot \frac{12}{\pi}\cos\frac{\pi t}{12} \right]_0^{12}$$

$$= \frac{1}{12} \left[ 60t - \frac{120}{\pi}\cos\frac{\pi t}{12} \right]_0^{12}$$

Evaluate at bounds:

At $t = 12$: $60(12) - \frac{120}{\pi}\cos(\pi) = 720 + \frac{120}{\pi}$

At $t = 0$: $0 - \frac{120}{\pi}\cos(0) = -\frac{120}{\pi}$

$$T_{\text{avg}} = \frac{1}{12}\left(720 + \frac{120}{\pi} + \frac{120}{\pi}\right)$$

$$= \frac{1}{12}\left(720 + \frac{240}{\pi}\right)$$

$$= 60 + \frac{20}{\pi}$$

$$\boxed{T_{\text{avg}} = 60 + \frac{20}{\pi} \approx 66.4°F}$$

The Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, then there exists at least one value $c$ in $[a, b]$ such that:

$$f(c) = f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx$$

This means the function actually attains its average value somewhere on the interval!

Example 5: Finding Where the Average Occurs

Problem: For $f(x) = x^2$ on $[0, 3]$, find the value $c$ where $f(c) = f_{\text{avg}}$.

From Example 2, we found $f_{\text{avg}} = 3$.

Set $f(c) = 3$:

$$c^2 = 3$$

$$c = \sqrt{3} \approx 1.732$$

$$\boxed{c = \sqrt{3}}$$

Since $\sqrt{3} \approx 1.732$ is in $[0, 3]$, this confirms the MVT for Integrals. ✓

Example 6: Average Value with Absolute Value

Problem: Find the average value of $f(x) = |x|$ on $[-2, 4]$.

Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \geq 0$, split the integral:

$$f_{\text{avg}} = \frac{1}{4-(-2)} \int_{-2}^{4} |x| \, dx$$

$$= \frac{1}{6} \left( \int_{-2}^{0} (-x) \, dx + \int_{0}^{4} x \, dx \right)$$

$$= \frac{1}{6} \left( \left[-\frac{x^2}{2}\right]_{-2}^{0} + \left[\frac{x^2}{2}\right]_{0}^{4} \right)$$

$$= \frac{1}{6} \left( (0 - (-2)) + (8 - 0) \right)$$

$$= \frac{1}{6}(2 + 8) = \frac{10}{6} = \frac{5}{3}$$

$$\boxed{f_{\text{avg}} = \frac{5}{3}}$$

Common Mistakes and Misunderstandings

❌ Mistake: Forgetting to divide by the interval length

Wrong: $f_{\text{avg}} = \int_a^b f(x) \, dx$

Why it's wrong: The integral alone gives the total "accumulation," not the average. You must divide by the interval length.

Correct: $f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx$

❌ Mistake: Using the wrong interval length

Wrong: For $f(x)$ on $[2, 5]$, using $\frac{1}{5}$ instead of $\frac{1}{3}$.

Why it's wrong: The denominator is $b - a$, not $b$.

Correct: $f_{\text{avg}} = \frac{1}{5-2} \int_2^5 f(x) \, dx = \frac{1}{3} \int_2^5 f(x) \, dx$

❌ Mistake: Thinking average = midpoint value

Wrong: Assuming $f_{\text{avg}}$ on $[a, b]$ equals $f\left(\frac{a+b}{2}\right)$.

Why it's wrong: This only works for linear functions! For curved functions, the average depends on how the function behaves across the entire interval.

Correct: Use the formula. For $f(x) = x^2$ on $[0, 3]$, the midpoint value is $f(1.5) = 2.25$, but the average is $3$.

❌ Mistake: Confusing average value with average rate of change

Wrong: Mixing up $f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$ with $\frac{f(b)-f(a)}{b-a}$.

Why it's wrong: Average value uses integration. Average rate of change uses the slope formula.

Correct:

Average value: $\frac{1}{b-a}\int_a^b f(x)\,dx$ (height of equivalent rectangle)
Average rate of change: $\frac{f(b)-f(a)}{b-a}$ (slope of secant line)

Average Function Value for MATH 122

Exam Relevance for MATH 122

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