Finding the Average Value of a Function on an Interval AB & BC

The average value of a continuous $f$ on $[a, b]$ is $\frac{1}{b - a}\int_a^b f(x)\,dx$: the integral is an area and the average value is the height of the rectangle on the same base enclosing that area. The four ways it goes wrong are all wrong divisors, and it is worth being able to name them: reporting the integral itself, dividing by $b$ instead of by $b - a$, averaging the two endpoint values, and evaluating $f$ at the midpoint of the interval. For $f(x) = x^{2}$ on $[1, 4]$ those produce $21$, $5.25$, $8.5$ and $6.25$ against a true average value of $7$.

The Mean Value Theorem for Integrals says a continuous function attains its average value: some $c$ in $[a, b]$ has $f(c) = f_{\text{avg}}$, so the curve crosses the top of the equal-area rectangle. That $c$ is an input and is rarely the midpoint; for $f(x) = x^{2}$ on $[0, 3]$ the average value is $3$ and $c = \sqrt{3}$. An average value carries the units of $f$ rather than of the integral, so the average of a rate is a rate, and multiplying it by $b - a$ recovers the total. Negative average values are ordinary and need no repair.