Monotonicity and Triangle Inequality for Integrals

Monotonicity and the triangle inequality for integrals are the quiet workhorses behind an enormous range of later results. The Mean Value Theorem for Integrals uses exactly the sup/inf bound of part (c) to trap the average value of a continuous function between its extremes, then invokes the Intermediate Value Theorem to locate a point where the average is achieved exactly. The Integral Test for Series compares a series to an integral by bounding rectangles above and below a decreasing function, a direct application of order preservation applied term by term. In probability and statistics, monotonicity of the integral underlies the fact that expectations preserve order: if a random variable $X \le Y$ almost surely, then $\mathbb{E}[X] \le \mathbb{E}[Y]$, a statement whose proof (in the measure-theoretic setting) mirrors the Riemann-sum argument given here almost verbatim. In numerical analysis, error bounds for quadrature rules (trapezoidal rule, Simpson's rule) rely on bounding $\left|\int f - Q(f)\right|$ using the triangle inequality applied to the difference of $f$ and its approximation. In physics and engineering, energy and norm estimates in differential equations frequently reduce to showing $\left|\int f \, dx\right| \le \int |f| \, dx \le \|f\|_\infty (b-a)$, chaining the triangle inequality with the sup/inf bound. In functional analysis and cryptographic protocol analysis (where integral or sum bounds control probability of failure events), the same two-line argument — replace a signed quantity by its absolute value, then bound by a supremum times a measure — recurs constantly, making this one of the most-reused proof patterns in all of analysis.

:::quiz[If $f(x) = -3$ for all $x \in [0,4]$, what does part (c) of the theorem give as a valid lower bound for $\int_0^4 f\,dx$, using $m=-3, M=-3$?]

• 0
• -3
• -12 ✓
• 12

:::quiz[Which hypothesis is essential for concluding $\int_a^b f \le \int_a^b g$ from $f(x)\le g(x)$ on $[a,b]$?]

• f and g must be differentiable
• f and g must both be Riemann integrable on [a,b] ✓
• f and g must be equal at the endpoints
• f and g must both be nonnegative

:::quiz[If $|f|$ is Riemann integrable on $[a,b]$, does it follow that $f$ is Riemann integrable?]

• Yes, always, by the triangle inequality
• No — |f| integrable does not imply f integrable, e.g. f = ±1 based on rationality ✓
• Yes, but only if f is continuous
• No, integrability of |f| implies f is unbounded

:::faq[Why is proving $|f|\in\mathcal R[a,b]$ a separate step rather than something obvious?] Because Riemann integrability is a nontrivial property tied to the oscillation of a function across partitions, and applying an operation to $f$ (like taking absolute value) could in principle worsen its oscillation. The proof must verify that it does not: using the reverse triangle inequality, one shows the oscillation of $|f|$ on any subinterval is never larger than the oscillation of $f$ itself. This lets a partition that makes $f$'s upper-minus-lower sum small automatically make $|f|$'s upper-minus-lower sum small too, which is precisely the Riemann integrability criterion.

Ask the tutor: Why must $|f|$'s integrability be proved separately instead of just assumed once $f$ is known integrable?

Picture the integral as the net signed area trapped between the graph of $f$ and the $x$-axis. If a second function $g$ never dips below $f$, then any rectangle you use to approximate the area under $g$ from below must be at least as tall as the corresponding rectangle for $f$. Every lower Darboux sum for $g$ dominates the corresponding lower sum for $f$, and the same argument works for upper sums. Since the integral is squeezed between these sums, it inherits the inequality. The triangle inequality is a sharper version of the same idea: no matter how much $f$ oscillates between positive and negative values, replacing $f$ by $|f|$ can only increase the total area being summed, because cancellation between positive and negative pieces of $f$ is destroyed once you take absolute values. The signed area of $f$ is thus always dominated in magnitude by the unsigned area of $|f|$.

Example 1 (order preservation with genuinely different functions). Let $f(x) = x^2$ and $g(x) = 2x + 3$ on $[0,2]$. We check $f(x) \le g(x)$ on this interval: the difference is

which is nonnegative exactly when $-1 \le x \le 3$; since $[0,2] \subset [-1,3]$, indeed $f(x) \le g(x)$ for all $x \in [0,2]$. Both $f$ and $g$ are continuous, hence Riemann integrable on $[0,2]$. Compute both integrals directly (using the Fundamental Theorem of Calculus, which presupposes only the integrability already established):

Indeed $\frac{8}{3} \approx 2.667 \le 10$, confirming Theorem part (b) numerically. Notice the functions are not close to each other in shape — one is a parabola, one is a line — which is the point: monotonicity of the integral requires nothing about the functions except the pointwise inequality and integrability. We can also illustrate part (c): on $[0,2]$, $f(x) = x^2$ satisfies $0 \le f(x) \le 4$ (with $m=0, M=4$), so the theorem predicts

which indeed holds, though the bound $8$ is far from tight — the sup/inf bound is a crude but always-valid estimate.

Example 2 (the triangle inequality with a sign-changing function). Let $f(x) = \sin(x) - \tfrac12$ on $[0, \pi]$, a function that is continuous (hence integrable) and changes sign on this interval, since $f(0) = -\tfrac12 < 0$ while $f(\pi/2) = \tfrac12 > 0$. First compute $\int_0^\pi f(x)\,dx$ directly:

Numerically, $\tfrac{\pi}{2} \approx 1.5708$, so $\int_0^\pi f \approx 2 - 1.5708 = 0.4292$, hence $\left|\int_0^\pi f\right| \approx 0.4292$. Now we need $\int_0^\pi |f(x)|\,dx = \int_0^\pi \left|\sin x - \tfrac12\right| dx$. The expression $\sin x - \tfrac12$ vanishes when $\sin x = \tfrac12$, i.e. at $x = \pi/6$ and $x = 5\pi/6$ in $[0,\pi]$. On $[0, \pi/6]$ and $[5\pi/6, \pi]$, $\sin x < \tfrac12$ so $f < 0$; on $[\pi/6, 5\pi/6]$, $f \ge 0$. Split the integral accordingly:

Using an antiderivative $F(x) = -\cos x - \tfrac{x}{2}$ for $\sin x - \tfrac12$ (so $-F(x) = \cos x + \tfrac{x}{2}$ is an antiderivative of $\tfrac12 - \sin x$): First piece:

Numerically: $\tfrac{\sqrt3}{2} \approx 0.8660$, $\tfrac{\pi}{12} \approx 0.2618$, sum $\approx 1.1278$, minus $1$ gives $\approx 0.1278$. Second piece:

Since $\cos(5\pi/6) = -\tfrac{\sqrt3}{2}$ and $\cos(\pi/6) = \tfrac{\sqrt3}{2}$:

Numerically: $\sqrt3 \approx 1.7321$, $\tfrac{\pi}{3}\approx 1.0472$, difference $\approx 0.6849$. Third piece (by the symmetry $x \mapsto \pi - x$ mapping $[5\pi/6,\pi]$ to $[0,\pi/6]$, and $\sin(\pi - x) = \sin x$): this equals the first piece, $\approx 0.1278$. Summing:

Check the theorem: $\left|\int_0^\pi f\right| \approx 0.4292 \le 0.9405 \approx \int_0^\pi |f|$. The inequality holds, and it is strict, exactly because $f$ changes sign — the negative contributions near the endpoints partially cancel the positive contribution in the middle when we do not take absolute values, so the signed integral is much smaller in magnitude than the unsigned one.