From a Rate to a Total Change
A rate describes how quickly a quantity changes. For example, velocity describes change in position per unit of time, and a flow rate describes change in volume per unit of time. Knowing a rate over an interval lets us ask how much the quantity changes across that whole interval.
If a rate stays constant, the calculation is familiar: multiply the rate by the length of time. If the rate varies, one multiplication is not enough. We can instead divide the interval into short pieces, estimate the change on each piece, and add those estimates. A definite integral formalizes the value approached as the pieces become finer.
Rate Times a Short Interval
Suppose \(r(t)\) is a rate at time \(t\). If the rate is approximately constant during a short time interval of length \(\Delta t\), then the change during that interval is approximately
This is the area of a rectangle whose height is the rate and whose width is the time interval. Its units are rate units multiplied by time units. For example, a rate in liters per minute multiplied by minutes gives liters.
Over a longer interval, rates may vary. Imagine dividing \([a,b]\) into \(n\) equal subintervals, each of width \(\Delta t=(b-a)/n\). Choose a time \(c_i\) in each subinterval and use \(r(c_i)\Delta t\) as an estimate of the change there. Adding the rectangle contributions gives an approximation to the total change.
For a continuous rate on a closed interval, these rectangular approximations approach a definite value as the widths shrink toward zero. That limiting value is the definite integral.
The symbols have useful roles: \(a\) and \(b\) are the interval endpoints, \(f(x)\) is the function being accumulated, and \(dx\) indicates the variable and the small interval widths involved. The definite integral is a number, not a function of \(x\).
Why Rectangular Slices Work
The key approximation is that a continuous rate changes only a limited amount over a sufficiently short interval. On a short slice, the rate is therefore close to its value at a chosen point in that slice. Multiplying that representative rate by the slice width estimates the change during the slice. Summing over the interval combines the estimated changes.
As the slices become narrower, the variation in the rate within each slice becomes smaller. For a continuous function on a closed interval, it is uniformly continuous: for any desired accuracy, sufficiently close inputs have function values as close as required. Thus, the difference between the rate on a slice and its representative value can be made small throughout every slice. The total error in the rectangle approximation then approaches zero as the largest slice width approaches zero. This is why the limit of the approximations gives the accumulated change.
Make short subintervals spanning the full interval from \(a\) to \(b\).
Use a representative rate multiplied by that subinterval's width.
The resulting rectangle total approximates the full accumulation.
The limit of the approximations is the definite integral.
Worked Example: Constant Rate
Worked Example: Constant Rate
A water tank gains water at a constant rate of \(5\) liters per minute for \(4\) minutes. Find the change in the tank's water volume.
Since the rate is constant, each minute contributes \(5\) liters. The total change is rate multiplied by elapsed time:
In integral notation, the same accumulation is \(\int_0^4 5\,dt=20\). Here every rectangular slice has height \(5\) liters per minute, so the sum of the slice contributions is exactly \(5\) times the total width \(4\) minutes.
Check: The result has units of liters, as a change in water volume should. Dividing \(20\) liters by \(4\) minutes recovers the stated rate of \(5\) liters per minute.
Worked Example: Estimating from Rate Data
Worked Example: Estimating from Rate Data
A vehicle's speed is recorded at the beginning of three consecutive one-hour intervals as \(40\), \(50\), and \(60\) miles per hour. Estimate the distance traveled during the three hours by treating each recorded speed as the rate for its hour.
For the first hour, the estimated distance is \(40\ \text{miles/hour}\times1\ \text{hour}=40\) miles. For the second hour, it is \(50\times1=50\) miles. For the third hour, it is \(60\times1=60\) miles. Adding the estimates gives
This is an estimate because the speed may change within each hour. In integral language, the desired distance is the accumulation of velocity over the three-hour interval, \(\int_0^3 v(t)\,dt\). The three rectangles approximate that integral using the recorded rates.
Check: The estimated average speed is \(150\ \text{miles}/3\ \text{hours}=50\) miles per hour, consistent with the three recorded values. The result is distance, not speed, because the hour units cancel in each rate-times-time product.
Worked Example: A Variable Rate and Its Limit
Worked Example: A Variable Rate and Its Limit
Suppose an object moves with velocity \(v(t)=2t\) meters per second from \(t=0\) to \(t=2\) seconds. Find the accumulated change in position by taking the limit of rectangular approximations that use the right end of each equal subinterval.
Divide the two-second interval into \(n\) equal pieces. Each width is \(\Delta t=2/n\) seconds. The right end of subinterval \(i\) is \(t_i=2i/n\), so its velocity is
The estimated change from that slice is its velocity times the width:
Adding these \(n\) contributions gives \(8/n^2\) times the sum of the integers from \(1\) through \(n\). That sum is \(n(n+1)/2\), so the total approximation is
As \(n\) increases without bound, \(4/n\) approaches zero. Therefore the rectangular approximations approach \(4\) meters:
Check: The average velocity over the interval is \(4\ \text{meters}/2\ \text{seconds}=2\) meters per second, matching the midpoint value \(v(1)=2\) meters per second for this linear velocity. Also, every approximation \(4+4/n\) is greater than \(4\), and it approaches \(4\) as the slices become finer.
Accumulation Has the Units of Change
When an integral accumulates a rate, its units are the product of the rate units and the input-variable units. If \(r(t)\) is measured in gallons per hour and \(t\) in hours, then \(\int_a^b r(t)\,dt\) is measured in gallons. If a position's velocity is measured in meters per second and time in seconds, the accumulated change in position is measured in meters.
This unit check is a useful way to interpret an answer. The rate itself is not the accumulated change. The definite integral combines the rate with the interval over which it acts.
A Common Mistake: Reporting the Rate Instead of the Change
Suppose a pump runs for \(3\) minutes at a constant rate of \(8\) liters per minute. A common error is to report \(8\) liters as the amount delivered. That number is the rate, with units of liters per minute, not the amount.
The correction is to multiply the rate by the elapsed time:
The units show why the incorrect calculation fails: liters per minute cannot be relabeled as liters, while multiplying by minutes cancels the time unit and gives the accumulated volume.
Why Definite Integrals Matter
Many quantities are naturally described by rates, while the questions we want to answer concern totals or net changes. A definite integral provides a single framework for connecting them: approximate many small changes, add them, and take the limit as the slices become arbitrarily narrow.
The same reasoning applies beyond motion and flow. If \(f(x)\) gives a changing rate of production, the integral over an interval accumulates the production. If it gives the rate at which a quantity is added to or removed from a system, its integral represents the resulting net change. Understanding this accumulation meaning gives the notation \(\int_a^b f(x)\,dx\) a concrete interpretation rather than treating it as a symbol to manipulate.
Check Your Understanding
For each question, focus on what is accumulating and on the units of the result.
- A leak drains water at a constant rate of \(3\) liters per minute for \(5\) minutes. What change in water volume does this rate produce?
- A rate is measured in kilograms per hour and integrated over an interval measured in hours. What are the units of the accumulated change?
- In a rectangular approximation, what two quantities are multiplied to estimate the change on one short interval?
- Why does making the rectangular slices narrower improve the approximation for a continuous rate?
- For \(v(t)=t\) meters per second on \(0\le t\le2\), write the definite integral that represents accumulated change in position. Do not evaluate it.
