From Throughput to Total Output
A production system may operate at different speeds at different times. A machine could run slowly while being set up, reach a high rate during steady operation, and slow again before stopping. A single rate value cannot describe all the output over the full period. To find the total, we need to account for how the production rate varies with time.
Calculus provides a precise model: the accumulated output over a time interval is the integral of the production rate on that interval. On a graph of rate against time, this integral is the area between the rate curve and the time axis. If production has already occurred before the interval begins, that starting amount must also be included when asking for the total output at a later time.
What the Rate and the Integral Mean
Let \(t\) measure time, and let \(p(t)\) denote the instantaneous production rate at time \(t\). For example, if time is measured in hours and output is measured in cartons, then \(p(t)\) has units of cartons per hour. The total output accumulated over a time interval has units of cartons.
The units follow directly from the integral. The rate contributes cartons per hour, and the time element contributes hours. Their product has units of cartons:
For a production process, throughput is ordinarily nonnegative: \(p(t)\geq 0\). A zero rate means production is stopped at that instant; a positive rate means output is being added. When the rate is nonnegative, the integral over an interval is also the ordinary geometric area under its graph.
The variable \(s\) in the integral is a dummy variable of integration. It represents time values ranging from the starting time \(a\) to the time \(t\). The output \(Q(t)\) is a total, while \(p(t)\) is a rate. Keeping those roles separate helps prevent mistakes with both interpretation and units.
Why Integration Gives Cumulative Production
The integral model comes from adding the output produced during many short time intervals. Suppose \(p\) is continuous on \([a,b]\). Divide this interval into short pieces. On a sufficiently short piece, the rate changes only slightly, so the output on that piece is approximately the rate at a selected time multiplied by the piece’s duration. Adding those products approximates the output over the whole interval.
As the pieces become shorter and more numerous, these sums converge to the definite integral. The continuity assumption ensures the rate is integrable and that the Fundamental Theorem of Calculus applies. The theorem then also shows that the instantaneous rate is the derivative of cumulative output.
For one such interval, let its duration be \(\Delta t_i\), and choose a time \(t_i^*\) inside it.
If the rate is nearly constant over that short interval, its output is approximately \(p(t_i^*)\Delta t_i\).
The sum \(\sum_i p(t_i^*)\Delta t_i\) approximates production from \(a\) to \(b\).
By the definition of the definite integral, the limiting sum is \(\int_a^b p(s)\,ds\).
Consequently, if \(Q(a)\) is the starting output, then \(Q(t)=Q(a)+\int_a^t p(s)\,ds\). For continuous \(p\), the Fundamental Theorem of Calculus gives
This derivative relation confirms the model’s meaning: the rate at which cumulative output changes at time \(t\) is the production rate at that time. It also gives a useful check on any formula for \(Q(t)\): differentiating it should recover \(p(t)\).
Worked Examples
Worked Example: A Polynomial Production Rate
A factory’s throughput during the first three hours is \(p(t)=80+12t+2t^2\) cartons per hour, where \(t\) is measured in hours. Assume production is zero at \(t=0\). Find the total output after three hours.
Use the cumulative-production formula with \(Q(0)=0\):
An antiderivative is \(80t+6t^2+\frac{2}{3}t^3\). Evaluating at the upper and lower endpoints gives
Check the rate by differentiating the accumulated-output formula \(Q(t)=80t+6t^2+\frac{2}{3}t^3\):
The derivative matches the given throughput, and \(Q(0)=0\), so both the rate and starting condition agree with the model.
Worked Example: Including Output Already Produced
At time \(t=0\), a production line has already made 250 cartons. From then on, its rate is \(p(t)=100+20t\) cartons per hour. Find the total output at hour 3 and the additional output made from hour 3 to hour 5.
First find the cumulative-output function, including the starting total:
At hour 3, substitute \(t=3\):
To find the additional output between hours 3 and 5, integrate over that interval. The initial 250 cartons are not added again, because this calculation asks only for production during the interval:
Check by evaluating the cumulative total at hour 5 and subtracting the total at hour 3:
The interval integral and the change in cumulative output agree.
Worked Example: A Sinusoidally Varying Rate
Suppose a line produces at rate \(p(t)=50+10\sin\left(\frac{\pi t}{6}\right)\) cartons per hour during the first six hours, with \(Q(0)=0\). Find its total output at hour 6.
The rate is nonnegative throughout the interval because the sine term ranges from 0 to 1 for \(0\leq t\leq 6\). Integrate the rate:
The decimal is rounded to two places only at the final step. For a check, the cumulative-output function is \(Q(t)=50t+\frac{60}{\pi}\left(1-\cos\left(\frac{\pi t}{6}\right)\right)\). Differentiating gives
Also, substituting \(t=0\) into \(Q(t)\) gives \(Q(0)=0\), as required.
Reading the Rate Graph
On a graph with time on the horizontal axis and production rate on the vertical axis, the integral from \(a\) to \(b\) measures the area under the rate curve between those times. For nonnegative throughput, this area represents the cartons, units, or other products made during that interval. A higher rate over some part of the interval contributes more output there; a longer interval at a given rate also contributes more output.
The graph’s vertical values are rates, not accumulated totals. For instance, a height of 120 cartons per hour at one time tells you the instantaneous throughput at that time. It does not say that 120 cartons have been produced since the beginning. To obtain total output, the rate must be accumulated over a duration by integration, with any starting total added separately.
When the rate is continuous, the Fundamental Theorem gives \(Q'(t)=p(t)\). Thus the slope of the cumulative-output function at a given time is the rate at that time. This is a consistency relationship between the two quantities; it does not make their units interchangeable. The slope has units of cartons per hour, while \(Q(t)\) has units of cartons.
A Common Mistake: Using the Final Rate for the Whole Period
For a variable rate, multiplying the rate at the endpoint by the whole elapsed time generally gives the wrong total. In the polynomial example, the rate at hour 3 is
A tempting but incorrect calculation is \(134\times 3=402\) cartons. This treats the line as if it had produced at the hour-3 rate throughout all three hours. But earlier rates were lower, so this calculation overstates the output. The correct calculation accounts for the changing rate at every time:
Multiplying one rate by a duration is valid when the rate is constant over that duration. Otherwise, use the integral of the rate. The relevant distinction is not whether a rate value is available, but whether that value represents the rate throughout the entire interval.
Choosing the Right Total
Before calculating, decide which quantity the question requests. “Output during the next two hours” asks for an interval integral. “Total output by hour 5” asks for the starting output plus the integral from the chosen baseline time to hour 5. If output is counted from the beginning of the model with \(Q(a)=0\), the cumulative total is just the integral from \(a\). If the system already has output at \(a\), omitting \(Q(a)\) gives only the new production, not the total.
Three checks make the model easier to interpret:
- Units: rate multiplied by time must produce output units.
- Starting value: the cumulative formula must return the stated baseline at its starting time.
- Rate check: when the rate is continuous, differentiating the cumulative formula should recover \(p(t)\).
These checks do not replace the integral calculation, but they can reveal common modeling errors such as a missing starting amount, an incorrect interval, or a formula with incompatible units.
Check Your Understanding
Use the rate, time interval, and starting output carefully in each question.
- A line has rate \(p(t)=30+5t\) units per hour on \(0\leq t\leq 4\), and begins with zero output. What is the cumulative output at hour 4?
- If \(p(t)\) is measured in kilograms per minute, what units does \(\int_a^b p(t)\,dt\) have when \(a\) and \(b\) are measured in minutes?
- A process has \(Q(2)=500\) items. Write an integral expression for its total output at time \(t=7\), using production rate \(p(t)\).
- In words, what does the area under a nonnegative rate graph from \(t=a\) to \(t=b\) represent?
- Why is multiplying \(p(b)\) by \(b-a\) not generally a valid way to find the output during \([a,b]\)?
