Accumulation and the Definite Integral

A pump’s flow rate is measured in litres per minute. Multiplying a constant rate by time gives transferred volume. For a changing rate, dividing the time interval into short pieces gives a sum of approximate transfers. The definite integral is the limit that makes this accumulation independent of the partition and sample choices, when such a limit exists.

Partitions and tagged sums

A partition $a=x_0<x_1<\cdots<x_n=b$ divides $[a,b]$ into widths $\Delta x_i=x_i-x_{i-1}$. Choose a tag $\xi_i$ in each subinterval. The Riemann sum is

$$\sum_{i=1}^n f(\xi_i)\Delta x_i.$$

The mesh is $\max_i\Delta x_i$. A bounded function is Riemann integrable when there is a real number $I$ such that every tagged sum is within any requested tolerance of $I$ once the mesh is sufficiently small. We denote that number by $\int_a^b f(x)\,dx$. The definition quantifies over all sufficiently fine partitions and tags; convergence of one favored sequence of sums is not by itself the full definition.

Continuous functions on closed bounded intervals are integrable. Bounded monotone functions are also integrable, even if they have jumps. Not every bounded function is: the function equal to one at rational inputs and zero at irrational inputs on $[0,1]$ admits tagged sums identically one or identically zero at every mesh, because both types of inputs occur in every subinterval. No common limit exists.

A square integrated from sums

For $f(x)=x^2$ on $[0,1]$, use $n$ equal pieces and right endpoints $i/n$. The sum is

$$R_n=\frac1n\sum_{i=1}^n\left(\frac in\right)^2 =\frac{n(n+1)(2n+1)}{6n^3} =\frac13+\frac1{2n}+\frac1{6n^2}.$$

We used $\sum i^2=n(n+1)(2n+1)/6$, which can be proved by induction. Thus $R_n\to1/3$. The left sum is $L_n=1/3-1/(2n)+1/(6n^2)$ and tends to the same limit. Because $x^2$ is increasing, every tagged sum for this equal partition lies between these endpoint sums; more generally, continuity supplies integrability for arbitrary fine partitions. Hence $\int_0^1x^2\,dx=1/3$.

The sums give a finite-error statement too: $R_n-L_n=1/n$. At $n=10$, the left and right values are $0.285$ and $0.385$. Both bracket $1/3$; neither is the exact integral. A picture of ten rectangles can illustrate the approximation without converting it into equality.

Signed accumulation and units

If $f$ is negative on part of the interval, its rectangular contributions are negative. Therefore the integral represents signed accumulation. Geometric area between a graph and the axis is $\int_a^b|f(x)|\,dx$, not necessarily $\int_a^bf(x)\,dx$.

For $f(x)=x$ on $[-1,1]$, symmetry gives signed integral zero, although geometric area is one. If $f$ is velocity, the integral gives displacement; integrating $|f|$ gives distance. If $f$ is flow in litres per minute and $x$ is minutes, the integral has units litres. An integral’s units multiply the integrand’s units by those of its input variable.

Properties from sums

Integrals are linear, additive across adjacent intervals, and preserve inequalities for integrable functions. Define $\int_b^af=-\int_a^bf$ and $\int_a^af=0$. The reversed orientation is a sign convention consistent with additivity, not a negative geometric interval length in the original partition definition.

If $m\leq f\leq M$ on $[a,b]$, then $m(b-a)\leq\int_a^bf\leq M(b-a)$. Also $|\int f|\leq\int|f|$. These bounds can detect an impossible answer before an exact calculation is available.

Exercises

1. Write the right-endpoint sum for $f(x)=3x$ on $[0,2]$ and calculate its limit.
2. Compute left and right sums for $x^2$ on $[0,1]$ with $n=4$.
3. If $2\leq f(x)\leq5$ on $[1,4]$, bound its integral.
4. Explain why an integral of a velocity can be zero while distance travelled is positive.
5. Show that changing a bounded integrable function at one point does not alter its integral.
6. Why does the rational-indicator example not become integrable when a single fine regular grid happens to choose only irrational tags?

Worked solutions

1. The width is $2/n$, tag $2i/n$, giving $(12/n^2)\sum i=6(n+1)/n\to6$.
2. $L_4=0.21875$ and $R_4=0.46875$, using the displayed formulas.
3. Between $6$ and $15$.
4. Opposite-direction displacements can cancel. Distance adds their magnitudes.
5. If the changed value differs by at most $B$, only subintervals touching that point can be affected. Their total contribution difference is bounded by $2B$ times the mesh, tending to zero. Interior tags usually affect one piece; allowing two handles shared endpoints.
6. Integrability requires all sufficiently fine tagged sums to converge to the same number. Rational tags still give one, while irrational tags give zero.