Additivity of the Integral

Suppose you want to integrate a function defined by two different formulas on two pieces of an interval — a ramp that flattens halfway through, say. The integral over the whole interval is just the sum of the integrals over the two pieces. This is additivity, and it turns out to be both easy to prove from first principles and far-reaching in its consequences: it underlies the variable-upper-limit function that connects the integral to differentiation.

Why additivity matters

Many functions in practice are piecewise-defined: they obey different formulas on different sub-intervals. Without additivity you would need a single formula to integrate over the entire domain; with it, you can break the domain at every junction point, integrate each piece, and add the results. Additivity also lets you subtract integrals: $\int_c^b f = \int_a^b f - \int_a^c f$, which is essential for estimating tails and for isolating contributions from sub-regions.

Formal statement and proof

Theorem (Additivity). Let $f$ be Riemann integrable on $[a, b]$, and let $c \in (a, b)$. Then $f$ is integrable on both $[a, c]$ and $[c, b]$, and

$$\int_a^b f(x)\,dx \;=\; \int_a^c f(x)\,dx \;+\; \int_c^b f(x)\,dx.$$

Proof. Recall that $f$ is integrable on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a partition $P$ of $[a, b]$ with $U(f, P) - L(f, P) < \varepsilon$, where $U$ and $L$ denote the upper and lower Darboux sums.

Step 1 — refining a partition by inserting $c$. Let $P$ be any partition of $[a, b]$, and let $P^* = P \cup \{c\}$ be the partition obtained by inserting the point $c$. If $c$ already belongs to $P$ then $P^* = P$. Otherwise $c$ falls in the interior of some sub-interval $[x_{k-1}, x_k]$ of $P$, which gets split into $[x_{k-1}, c]$ and $[c, x_k]$. For the upper sum, the supremum over the original interval satisfies

$$M_k \;=\; \sup_{[x_{k-1},x_k]} f \;\geq\; \sup_{[x_{k-1},c]} f \quad\text{and}\quad M_k \;\geq\; \sup_{[c,x_k]} f,$$

so replacing $M_k(x_k - x_{k-1})$ with two smaller (or equal) terms can only decrease the upper sum: $U(f, P^*) \leq U(f, P)$. By the analogous argument for infima, $L(f, P^*) \geq L(f, P)$. Therefore

$$U(f, P^*) - L(f, P^*) \;\leq\; U(f, P) - L(f, P).$$

Step 2 — integrability on sub-intervals. Since $f$ is integrable on $[a, b]$, for any $\varepsilon > 0$ pick $P$ with $U(f,P) - L(f,P) < \varepsilon$. Refine to $P^*$ as above, and set $P_1 \coloneqq P^* \cap [a, c]$ and $P_2 \coloneqq P^* \cap [c, b]$. The Darboux sums split:

$$U(f, P^*) = U(f, P_1) + U(f, P_2), \qquad L(f, P^*) = L(f, P_1) + L(f, P_2).$$

Since each term is non-negative,

$$U(f, P_1) - L(f, P_1) \;\leq\; U(f, P^*) - L(f, P^*) < \varepsilon,$$

and the same holds for $P_2$. As $\varepsilon > 0$ was arbitrary, $f$ is integrable on $[a, c]$ and on $[c, b]$.

Step 3 — the sum formula. Denote $I \coloneqq \int_a^b f$, $I_1 \coloneqq \int_a^c f$, $I_2 \coloneqq \int_c^b f$. For any partition $P^*$ of $[a,b]$ containing $c$:

$$L(f, P_1) + L(f, P_2) \;\leq\; I_1 + I_2 \;\leq\; U(f, P_1) + U(f, P_2).$$

But the outer expressions equal $L(f, P^*)$ and $U(f, P^*)$ respectively, and both converge to $I$ as the mesh of $P^*$ shrinks. Hence $I_1 + I_2 = I$. $\square$

Orientation convention and signed integrals

The Darboux definition implicitly assumes $a < b$. To handle all orderings uniformly, adopt the conventions

$$\int_a^a f(x)\,dx \;\coloneqq\; 0, \qquad \int_b^a f(x)\,dx \;\coloneqq\; -\int_a^b f(x)\,dx \quad (a < b).$$

With these in place, the additivity formula $\int_a^b f = \int_a^c f + \int_c^b f$ holds for any ordering of $a$, $b$, $c$ on the real line. To see why, consider the case $c < a < b$:

$$\int_a^b f \;=\; \int_c^b f - \int_c^a f \;=\; \int_c^b f + \int_a^c f,$$

which is just the original additivity formula with the roles of $c$ and $a$ swapped. All other cases follow similarly. The sign conventions ensure you can freely insert or remove intermediate limits without tracking which endpoint is larger.

The variable-upper-limit function

Fix a base point $a$ in the domain of integration, and define

$$F(x) \;\coloneqq\; \int_a^x f(t)\,dt$$

for all $x$ in the interval $[a, b]$. This is the variable-upper-limit function of $f$ based at $a$ (also called the integral function).

Well-definedness. By additivity applied to $[a, x] \subset [a, b]$, integrability of $f$ on $[a, b]$ implies integrability on $[a, x]$ for every $x \in [a, b]$. Hence $F(x)$ is a well-defined real number for each such $x$.

Change of base point. If you replace $a$ by another base point $a'$, the resulting function $F_{a'}(x) = \int_{a'}^x f(t)\,dt$ satisfies

$$F_{a'}(x) \;=\; F(x) \;-\; \underbrace{F(a')}_{\text{constant}} \;=\; F(x) + C.$$

Changing the base point shifts $F$ by a constant. This is the first hint of why primitives are unique only up to an additive constant — a theme developed fully in the Primitives checkpoint.

The function $F$ is the crucial bridge to the Newton–Leibniz formula: when $f$ is continuous, $F$ is differentiable with $F'(x) = f(x)$, which means $F$ is a primitive of $f$.

Worked example: piecewise linear function

Consider the tent function

$$f(x) \;=\; \begin{cases} x & 0 \leq x \leq 1, \\ 2 - x & 1 < x \leq 2. \end{cases}$$

It rises linearly from $0$ to $1$ on $[0,1]$, then falls back to $0$ on $[1,2]$. Apply additivity at $c = 1$:

$$\int_0^2 f(x)\,dx \;=\; \int_0^1 x\,dx \;+\; \int_1^2 (2-x)\,dx.$$

The graph of each piece is a right triangle with base $1$ and height $1$, so each integral equals $\tfrac{1}{2}(1)(1) = \tfrac{1}{2}$. Therefore

$$\int_0^2 f(x)\,dx \;=\; \frac{1}{2} + \frac{1}{2} \;=\; 1.$$

Now use the variable-upper-limit function to trace how the accumulated area grows. For $x \in [0,1]$,

$$F(x) = \int_0^x t\,dt = \frac{x^2}{2}.$$

For $x \in [1,2]$, split at $1$:

$$F(x) = \int_0^1 t\,dt + \int_1^x (2-t)\,dt = \frac{1}{2} + \left[2t - \frac{t^2}{2}\right]_1^x = \frac{1}{2} + 2x - \frac{x^2}{2} - \frac{3}{2} = 2x - \frac{x^2}{2} - 1.$$

You can check that $F(0) = 0$, $F(1) = \tfrac{1}{2}$, and $F(2) = 1$, consistent with the areas computed above.

Summary

• Additivity: if $f$ is integrable on $[a, b]$ and $c \in (a, b)$, then $f$ is integrable on $[a, c]$ and $[c, b]$, and $\int_a^b f = \int_a^c f + \int_c^b f$.
• Proof: insert $c$ into any partition (this cannot increase $U - L$), split sums exactly at $c$, then pass to the limit.
• Orientation conventions: $\int_a^a f \coloneqq 0$ and $\int_b^a f \coloneqq -\int_a^b f$ make additivity valid for all orderings of $a, b, c$.
• Variable-upper-limit function: $F(x) \coloneqq \int_a^x f(t)\,dt$ is well-defined on $[a, b]$; changing the base point shifts $F$ by a constant.
• $F$ is the bridge to the Newton–Leibniz formula: for continuous $f$, $F'(x) = f(x)$.