The height with the same area AB & BC

For $f$ continuous on $[a, b]$, the average value is

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

Read the formula as a picture rather than as a rule. The integral is the area under the curve. Divide that area by the width $b - a$ and you get the height of the rectangle on the same base with the same area. That rectangle is what an average value is.

1. Integrate over the interval. This is the area, and it is not yet the answer.
2. Divide by $b - a$. The length of the interval, never an endpoint.
3. Report a value of $f$. If the number you wrote down could not plausibly be an output of $f$, something went wrong.

That last check catches a great deal. Take $f(x) = x^{2}$ on $[0, 3]$. The integral is $9$, but $f$ never gets above $9$ and spends most of the interval far below it, so $9$ cannot be the average of its own values. Dividing by $3$ gives $3$, which sits comfortably inside the range of $f$ on that interval.

The definition also explains why the discrete idea of an average survives: adding up $n$ sample values and dividing by $n$ becomes, in the limit, integrating and dividing by the length. The $\frac{1}{b-a}$ is the continuous version of the $\frac{1}{n}$, which is why forgetting it is not a small error.

Every common error in this topic is a wrong divisor, so it is worth naming all four. Take $f(x) = x^{2}$ on $[1, 4]$, where $\int_1^4 x^{2}\,dx = 21$ and the average value is $7$.

1. Dividing by nothing. Reporting $21$. This is the single most common answer and it is an area, not a height.
2. Dividing by $b$. Reporting $\frac{21}{4} = 5.25$. The formula asks for the length of the interval, and $[1, 4]$ has length $3$, not $4$. Intervals that start at $0$ hide this error completely, which is why it survives until an interval that does not.
3. Averaging the endpoint values. Reporting $\frac{f(1) + f(4)}{2} = \frac{1 + 16}{2} = 8.5$. This uses two of the infinitely many values of $f$ and ignores the rest.
4. Evaluating at the midpoint of the interval. Reporting $f(2.5) = 6.25$. This averages the inputs and then applies $f$. The average of $f$ is not $f$ of the average, and the two agree only for linear functions.

Two of those four give answers close to $7$, which is the reason they persist. Being close is not a check; dividing by the length is.

An average value is a number. The theorem says that a continuous function actually attains it.

Mean Value Theorem for Integrals. If $f$ is continuous on $[a, b]$, then there is at least one $c$ in $[a, b]$ with $f(c) = f_{\text{avg}}$, that is, with $\int_a^b f(x)\,dx = f(c)(b - a)$.

Geometrically: somewhere on the interval the curve passes through the top edge of the equal-area rectangle. For $f(x) = x^{2}$ on $[0, 3]$ the average value is $3$, and solving $c^{2} = 3$ gives $c = \sqrt{3} \approx 1.732$, which is indeed inside $[0, 3]$.

Three things to keep straight about $c$:

1. $c$ is an input, $f_{\text{avg}}$ is an output. Setting $c$ equal to the average value is a units error dressed as algebra.
2. $c$ is rarely the midpoint. Here the midpoint is $1.5$ and $c$ is about $1.732$. They coincide when $f$ is linear and generally not otherwise.
3. Continuity is the hypothesis. A function with a jump can have an average value it never takes, in the same way the Intermediate Value Theorem fails without continuity.

Note the family resemblance to the Mean Value Theorem of Unit 5: that one produces a $c$ where the instantaneous rate equals the average rate, this one produces a $c$ where the value equals the average value. Both are existence statements, and neither locates $c$ for you.

An average value carries the units of $f$. It does not carry the units of the integral.

If $r(t)$ is measured in liters per minute and $t$ in minutes, then $\int_0^{12} r(t)\,dt$ is in liters, since the units multiply. Dividing by $12$ minutes puts you back in liters per minute. So the average value of a rate is a rate, and answering "liters" describes the integral instead.

That gives a fast check on the arithmetic. Whatever $f$ measures, the average value measures the same thing, on the same scale. A number that could not be a value of $f$ is not the average value of $f$.

The relation runs both ways, which is how these problems are usually posed on the exam. Since $\int_a^b f = f_{\text{avg}} \cdot (b - a)$, an average value plus an interval length gives a total. A tap running at an average of $6$ gallons per minute for $20$ minutes delivers $120$ gallons; the average alone is not an amount, and multiplying by the elapsed time is what turns it into one.

Finally, the sign. Nothing here assumes $f \geq 0$. On $[0, 3]$ the function $f(x) = x^{2} - 4$ has integral $-3$ and average value $-1$, and a negative average value is the correct report for a function that spends most of the interval below the axis. Taking an absolute value at the end answers a different question.