Once a function is integrable, simple inequalities about $f$ become inequalities about $\int f$. Combined with the extreme value theorem, they give the mean value theorem for integrals, a key step in the proof of the fundamental theorem.
Key idea
If $m \le f(x) \le M$ on $[a, b]$, then $m(b - a) \le \int_a^b f \le M(b - a)$. Also $\left|\int_a^b f\right| \le \int_a^b |f|$. Mean value theorem for integrals: if $f$ is continuous on $[a, b]$, then $f(c) = \frac{1}{b - a}\int_a^b f$ for some $c \in [a, b]$.
Why it works
For the one-piece partition $\{a, b\}$, the lower sum is at least $m(b - a)$ and the upper sum at most $M(b - a)$, and the integral sits between all lower and upper sums. For the absolute value, $-|f| \le f \le |f|$ integrates to $-\int |f| \le \int f \le \int |f|$. For the mean value theorem, let $m$ and $M$ be the minimum and maximum of $f$ (they exist by the EVT). The average $\frac{1}{b - a}\int_a^b f$ lies between $m$ and $M$, so by the intermediate value theorem $f$ takes that value at some $c$. $\blacksquare$
Worked examples
Worked example: Bounds without integrating
Bound $\int_2^6 \left(-x^{2} + 6x - 2\right) dx$ using the minimum and maximum of the integrand.
- The vertex is at $x = 3$, inside $[2, 6]$, with value $7$: the maximum.
- At the ends: $f(2) = 6$ and $f(6) = -2$; the minimum is $-2$.
Answer: $-8 \le \int_2^6 f \le 28$. (The true value is $\frac{56}{3}$.)
Worked example: The point of the mean value theorem
Find $c$ in $[0, 3]$ with $f(c)$ equal to the average of $f(x) = x^{2}$ on $[0, 3]$.
- The average is $\frac{1}{3}\int_0^3 x^{2}\,dx = \frac{1}{3} \cdot 9 = 3$.
- $c^{2} = 3$ with $c \ge 0$.
Answer: $c = \sqrt{3}$
Worked example: Cancellation
For $f(x) = x - 1$ on $[-2, 3]$, compute $\left|\int f\right|$ and $\int |f|$.
- $\int_{-2}^{3} (x - 1)\,dx = \left[\frac{x^{2}}{2} - x\right]_{-2}^{3} = \frac{3}{2} - 4 = -\frac{5}{2}$.
- For $\int |f|$, split at $x = 1$: $\int_{-2}^{1} (1 - x)\,dx = \frac{9}{2}$ and $\int_1^3 (x - 1)\,dx = 2$.
Answer: $\left|\int f\right| = \frac{5}{2}$, $\int |f| = \frac{13}{2}$.
Common mistake
Using the values at the ends as the bounds. The bounds need the true minimum and maximum of $f$ on the interval; for a function that turns, those are often inside. In the first example, the ends give $6$ and $-2$, but the maximum is $7$. Also keep $\left|\int f\right| \le \int |f|$ pointing the right way: positive and negative parts cancel inside $\int f$, never inside $\int |f|$.