Additivity Over Intervals

From Two Adjacent Intervals to a Finite Partition

The Order Preservation of the Integral tutorial used comparisons between integrals on a fixed interval. Here the focus is how an integral over one interval relates to integrals over its subintervals. The Additivity of the Riemann Integral Across Adjacent Intervals Theorem, established in Properties of the Riemann Integral, says that if $a<c<b$ and $f$ is Riemann integrable on $[a,b]$, then

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

The restriction of an integrable function to each of these subintervals is integrable by the Restriction to Subintervals Theorem. We will use these earlier results to establish a finite-partition formula: splitting an interval at several points gives a sum of all the subinterval integrals. The formula makes it possible to calculate piece by piece and to rearrange contributions from different parts of an interval.

Theorem (Finite Additivity Across a Partition): Let $f:[a,b]\to\mathbb{R}$ be Riemann integrable, and let $a=x_0<x_1<\cdots<x_n=b$ be a partition of $[a,b]$. Then $f$ is Riemann integrable on each $[x_{i-1},x_i]$, and $$ \int_a^b f(x)\,dx = \sum_{i=1}^{n}\int_{x_{i-1}}^{x_i} f(x)\,dx. $$

Proof. Integrability on each subinterval follows from the Restriction to Subintervals Theorem. We prove the integral identity by induction on the number $n$ of subintervals. If $n=1$, then $x_0=a$ and $x_1=b$, so the asserted identity is exactly $\int_a^b f=\int_a^b f$.

Suppose the identity holds for partitions with $n-1$ subintervals, where $n\geq2$. Since $a<x_{n-1}<b$, the Additivity of the Riemann Integral Across Adjacent Intervals Theorem gives

\int_a^b f(x)\,dx = \int_a^{x_{n-1}} f(x)\,dx + \int_{x_{n-1}}^b f(x)\,dx.

Apply the induction hypothesis to the partition $a=x_0<\cdots<x_{n-1}$ of $[a,x_{n-1}]$. It gives

\int_a^{x_{n-1}} f(x)\,dx = \sum_{i=1}^{n-1}\int_{x_{i-1}}^{x_i}f(x)\,dx.

Because $x_n=b$, the remaining term is $\int_{x_{n-1}}^b f=\int_{x_{n-1}}^{x_n}f$. Substituting these expressions into the two-interval identity proves the formula for $n$ subintervals. The induction is complete. $\square$

The result says that a finite subdivision does not change the total integral; it only expresses that total as a sum of local contributions. The terms are signed contributions: where $f$ is negative, the corresponding integral can be negative. Additivity does not mean that every piece is a positive area.

Calculating by Splitting at Convenient Points

A partition is especially useful when the integrand has a simpler expression on each subinterval, or when the integral on each piece is already known. Each calculation must use the same function restricted to that piece. The theorem then combines those local integrals without requiring a new calculation over the whole interval.

Worked Example: Splitting a Polynomial Integral into Three Pieces

Let $f(x)=x^2+1$ on $[0,3]$, and use the partition $0<1<2<3$. Finite additivity gives

\int_0^3(x^2+1)\,dx = \int_0^1(x^2+1)\,dx +\int_1^2(x^2+1)\,dx +\int_2^3(x^2+1)\,dx.

The three terms are

\begin{aligned} \int_0^1(x^2+1)\,dx&=\frac{1^3-0^3}{3}+(1-0)=\frac43,\\ \int_1^2(x^2+1)\,dx&=\frac{2^3-1^3}{3}+(2-1)=\frac73+1=\frac{10}{3},\\ \int_2^3(x^2+1)\,dx&=\frac{3^3-2^3}{3}+(3-2)=\frac{19}{3}+1=\frac{22}{3}. \end{aligned}

Adding the pieces yields

\int_0^3(x^2+1)\,dx = \frac43+\frac{10}{3}+\frac{22}{3} = \frac{36}{3} =12.

Direct evaluation gives $\frac{3^3}{3}+3=9+3=12$, in agreement. The split does not alter the value; it organizes the calculation into interval contributions.

Piecewise-Defined Functions

Breakpoints in a piecewise formula naturally specify a partition. The function need not have the same formula on both sides of a breakpoint. It must, however, be Riemann integrable on the full interval for finite additivity to apply. Step functions are integrable by the Integrability and Integral of a Step Function Theorem, and restrictions to the pieces remain integrable.

Worked Example: Adding the Contributions of a Step Function

Define $f:[0,4]\to\mathbb{R}$ by $f(x)=2$ for $0\leq x<1$, $f(x)=-1$ for $1\leq x<2$, and $f(x)=3$ for $2\leq x\leq4$. This is a step function and is therefore Riemann integrable. For the partition $0<1<2<4$, finite additivity gives

\int_0^4 f(x)\,dx = \int_0^1 f(x)\,dx+\int_1^2 f(x)\,dx+\int_2^4 f(x)\,dx.

On $[0,1]$, $f$ equals $2$ except at the endpoint $1$; changing a function at one point does not change its integral, by the Finite Point Changes Do Not Affect the Integral Theorem. Thus $\int_0^1 f(x)\,dx=2(1-0)=2$. Similarly, on $[1,2]$, it equals $-1$ except at the endpoint $2$, so $\int_1^2 f(x)\,dx=-1(2-1)=-1$. On $[2,4]$, it equals $3$ throughout, and its integral is $3(4-2)=6$. Consequently,

\int_0^4 f(x)\,dx=2+(-1)+6=7.

The middle contribution is negative. The final result is the signed sum of the contributions, not the sum of their absolute values. The values assigned at the finitely many breakpoints do not change the integral, though the function's integrability and the integral formula still need to be justified.

Recovering an Integral on One Part

Finite additivity can also be used in reverse. If the integral over a whole interval and the integral over one part are known, subtraction determines the integral over the remaining part. This is an algebraic consequence of the partition formula, not a separate assumption about the function's sign.

Worked Example: Finding an Unknown Subinterval Integral

Suppose $f$ is Riemann integrable on $[1,5]$, with $\int_1^5 f(x)\,dx=11$ and $\int_1^3 f(x)\,dx=4$. The partition $1<3<5$ gives

\int_1^5 f(x)\,dx = \int_1^3 f(x)\,dx+\int_3^5 f(x)\,dx.

Substituting the given values,

11=4+\int_3^5 f(x)\,dx, \qquad\text{so}\qquad \int_3^5 f(x)\,dx=7.

No pointwise sign condition is needed. If $f$ takes both positive and negative values on $[3,5]$, its integral is still determined by the same additive identity.

An Identity for Overlapping Intervals

Two intervals can overlap without one containing the other. The finite-partition formula lets us compare their integrals by dividing their union at every endpoint. In particular, the contributions on the overlap occur once in each of two sums and can be rearranged to give an identity.

Theorem (Integral Identity for Overlapping Intervals): Let $f$ be Riemann integrable on $[a,d]$, where $a<b<c<d$. Then $$ \int_a^c f(x)\,dx+\int_b^d f(x)\,dx = \int_a^d f(x)\,dx+\int_b^c f(x)\,dx. $$

Proof. The restrictions of $f$ to all subintervals in question are integrable. Apply finite additivity to the partitions $a<b<c$, $b<c<d$, and $a<b<c<d$, respectively. This gives

\begin{aligned} \int_a^c f&=\int_a^b f+\int_b^c f,\\ \int_b^d f&=\int_b^c f+\int_c^d f,\\ \int_a^d f&=\int_a^b f+\int_b^c f+\int_c^d f. \end{aligned}

Adding the first two equalities gives $\int_a^c f+\int_b^d f=\int_a^b f+2\int_b^c f+\int_c^d f$. Adding $\int_b^c f$ to the third equality gives the same expression on the right: $\int_a^d f+\int_b^c f=\int_a^b f+2\int_b^c f+\int_c^d f$. Therefore the two sides of the asserted identity are equal. $\square$

The identity is useful when integrals over overlapping intervals are known but the integral over their union or intersection is not. It is also a consistency check: each side counts the parts outside the overlap once and the overlap twice.

Worked Example: Checking an Overlap Identity

Take $f(x)=x$ on $[0,4]$, with $a=0$, $b=1$, $c=3$, and $d=4$. The theorem states that

\int_0^3 x\,dx+\int_1^4 x\,dx = \int_0^4 x\,dx+\int_1^3 x\,dx.

Evaluate each term:

\int_0^3x\,dx=\frac{9}{2},\qquad \int_1^4x\,dx=\frac{16-1}{2}=\frac{15}{2},

\int_0^4x\,dx=\frac{16}{2}=8,\qquad \int_1^3x\,dx=\frac{9-1}{2}=4.

The left side is $9/2+15/2=24/2=12$, and the right side is $8+4=12$. The shared interval $[1,3]$ is counted on both sides in exactly the way the identity describes.

What Additivity Does—and Does Not—Say

Additivity is a statement about integrals on intervals that fit together at endpoints. Shared endpoints do not create an extra contribution: a point has no interval length, and finite point changes do not affect the integral of an integrable function. This is why a piecewise formula can be split at its breakpoints without adding a correction for the endpoint values.

The formula also does not claim that the integral on each piece is positive, or that the integral records the maximum or minimum value of the function there. Each piece contributes its own signed integral. If one piece contributes a negative amount, it subtracts from the total. When using the theorem, verify that the pieces form the intended partition in order and that the function is integrable on the original interval. Integrability on the pieces then follows from restriction.

For practical use, a useful sequence is to mark all breakpoints, calculate the integral on each resulting subinterval, and add the results with their signs. If intervals overlap rather than form a partition, divide at all endpoints first or use the overlap identity. These methods rely on finite additivity and avoid treating overlapping contributions as though they were disjoint.

List the endpoints in increasing order.
They determine the partition and the consecutive subintervals.

Check integrability.
Use integrability on the full interval and restriction to obtain integrability on each piece.

Compute each signed contribution.
Keep negative integrals negative; add the resulting numbers.

For overlapping intervals, track the overlap.
Split at the endpoints and use the integral identity to avoid counting a piece incorrectly.

Check Your Understanding

Use finite additivity and the interval identities from this tutorial to answer the following questions.

State the finite-additivity formula for a partition $a=x_0<x_1<\cdots<x_n=b$.
If $\int_0^6 f=13$ and $\int_0^2 f=5$, what is $\int_2^6 f$?
Why does a negative integral on one subinterval not prevent applying finite additivity?
For $a<b<c<d$, write the identity relating the integrals on $[a,c]$, $[b,d]$, $[a,d]$, and $[b,c]$.
Why does changing the value of a piecewise-defined integrable function at a finite number of breakpoints not add a separate endpoint term to its integral?

What You'll Learn