Textbook
The function $f$ is continuous on a finite interval $[a,b]$, with $a<b$. A partition has points $a=t_0<\cdots<t_n=b$, widths $\Delta t_i=t_i-t_{i-1}$, and selected points $\xi_i\in[t_{i-1},t_i]$. The Riemann integral is the limit of weighted sums as the largest interval width tends to zero:
$$\int_a^b f(t)\,dt=\lim_{\max_i\Delta t_i\to0}\sum_{i=1}^{n}f(\xi_i)\Delta t_i.$$
The limit must be the same for every such partition and choice of sample points. Continuity guarantees its existence. Heights multiplied by widths are added, giving a total accumulated contribution over a continuous interval.
If $f(t)=2$ on $[0,3]$, the integral is 6. Negative function values contribute negatively: $\int_{-1}^{1}t\,dt=0$, even though the geometric area between graph and axis is positive. An area interpretation must account for signs.
The symbol $dt$ identifies the integration variable. For an integrable probability density $p_X$, an interval’s probability is $\int_a^b p_X(t)\,dt$ and an individual point has probability zero. Infinite bounds, unbounded functions, and more general integrals require separate definitions and condition checks.
The width factor $\Delta t_i$ is essential: summing heights alone would grow with the number of pieces even for a constant function. Refining a partition reduces widths while increasing the number of terms. Requiring the largest width to shrink prevents leaving one large piece untouched while refining the others. A continuous function on a finite closed interval varies little on sufficiently short pieces, so different admissible sample heights eventually make little difference to the sum.
Definitions and notation: Basic Analysis I.
Related concepts: Function limits: making “approaches” precise, Finite Sums: Reading the Summation Sign.
In plain language
Accumulated water volume can be approximated over short intervals: a representative flow rate is multiplied by each duration, and the contributions are added. Litres per minute multiplied by minutes give litres; merely adding meter readings exposes a mistake in the units alone. Pumping water out contributes negatively, so an integral cannot rebrand every withdrawal as income for a prettier report. Each interval temporarily replaces a changing flow with a constant one. The integral describes the limit of those sums under its stated conditions, not the accuracy of one convenient reading.
Example
For the constructed example $f(t)=t$ on $[0,1]$, the selected points are right endpoints of equal-width intervals. The sums approach $1/2$.
The Riemann integral: accumulating signed contributions
One interval
Right-endpoint height and width are both 1, giving sum 1.
Step 1 of 4: One interval
Read all steps
One interval
Right-endpoint height and width are both 1, giving sum 1.
- Contributions
- 1current
- Sum
- 1
Two intervals
Contributions 0.25 and 0.5 sum to 0.75.
- Contributions
- 0.25included0.5current
- Sum
- 0.75
Four intervals
A finer partition gives sum 0.625.
- Contributions
- 0.0625included0.125included0.1875included0.25current
- Sum
- 0.625
General expression
For n equal intervals, the sum is (n + 1)/(2n), whose limit is 1/2.
- Interval counts
- 410100
- Sums
- 0.6250.55current0.505included
Because $f(t)=t$ increases, each right endpoint supplies its piece’s greatest height, so the displayed sums exceed the integral. The final formula $(n+1)/(2n)=1/2+1/(2n)$ exposes the extra contribution, which disappears as $n$ grows. The four-piece value 0.625 is still a sum, not the integral. Treating the last displayed grid as exact merely because the small rectangles look convincing would confuse approximation with the limiting answer.
The example uses one choice of sample points. The definition requires the same limit under every admissible refinement.