From a Production Rate to a Total
A production rate describes how quickly units are being made at a particular time. A rate of \(80\) units per hour means that, near that time, production is increasing by about \(80\) units for each hour that passes. If the rate changes during a shift, multiplying one rate by the full shift length may not give the correct output.
Integration handles that change. It adds up the output produced over many short time intervals, using the rate appropriate to each interval. The result is the total number of units produced during the specified period.
Let \(r(t)\) be a production rate at time \(t\). If time is measured in hours, then \(r(t)\) might have units of units per hour. The product \(r(t)\,dt\) has units of units, so its integral over a time interval represents an amount of production.
The nonnegative-rate condition matches the usual meaning of production: the process makes units at a rate of zero or greater. If production stops during part of the interval, set \(r(t)=0\) there. That period contributes zero to the integral, but it remains part of the elapsed time.
Why the Integral Gives the Total
Over a short interval of duration \(\Delta t\), if the rate is approximately \(r(t)\), the output during that interval is approximately \(r(t)\Delta t\). For example, a rate near \(50\) units per hour over a short interval of \(0.1\) hour corresponds to about \(50(0.1)=5\) units.
Divide \([a,b]\) into short subintervals. Approximate the output on each by the rate at a sample time multiplied by the subinterval's duration, then add the approximations. As the subintervals become finer, these sums approach the definite integral, provided the rate is integrable. For a continuous rate, the Fundamental Theorem of Calculus gives a direct way to evaluate that integral.
The theorem connects the rate model to an accumulated amount. Indeed, define the production accumulated from time \(a\) up to time \(x\) by \(A(x)=\int_a^x r(t)\,dt\). Where \(r\) is continuous, the Fundamental Theorem of Calculus states that \(A'(x)=r(x)\). Thus the rate is the derivative of the accumulated production, and integrating recovers the change in production.
Worked Example: A Continuously Increasing Rate
Worked Example: A Continuously Increasing Rate
A machine produces at rate \(r(t)=120+18t\) units per hour, where \(t\) is the number of hours after the shift begins. Find the production during the first \(5\) hours.
The total is the integral of the rate from \(t=0\) to \(t=5\):
An antiderivative is \(F(t)=120t+9t^2\), since \(F'(t)=120+18t\). Therefore,
The machine produces \(825\) units in the first \(5\) hours. The units also check: units per hour integrated with respect to hours gives units. As a check, differentiating \(120t+9t^2\) returns \(120+18t\), the given rate.
Piecewise Schedules
Production schedules often use different operating rates in different parts of a shift. A piecewise-defined rate model gives a formula on each time interval. To find the total, integrate each piece over its own interval and add the results.
For adjacent intervals, an endpoint may appear as the end of one interval and the beginning of the next. This does not count production twice: changing the value of a rate at a single time does not change a definite integral. What matters is using each formula over the correct duration.
Worked Example: A Schedule with Two Operating Rates and a Stop
A workshop produces at \(60\) units per hour during the first \(2\) hours, at \(90\) units per hour during the next \(2\) hours, and then stops for \(1\) hour. Find its total production over the \(5\)-hour period.
With \(t\) measured in hours from the start, one rate model is
Integrate on all three intervals:
The workshop produces \(300\) units. The first interval contributes \(120\) units, the second contributes \(180\), and the stopped interval contributes none. The schedule check is consistent: the rate is zero throughout the final hour, so extending the end time from hour \(4\) to hour \(5\) adds no production.
Slowdowns and Stoppages
A slowdown is different from a complete stop. During a slowdown, the rate is positive but smaller; during a stop, it is zero. If the rate decreases continuously, its integral includes the output produced while that decrease is happening. It is not correct to treat the entire slowdown interval as either the initial rate or the final rate.
Worked Example: A Linear Slowdown Followed by a Stop
A production line runs at \(120\) units per hour for the first \(2\) hours. Over the next \(4\) hours, its rate decreases linearly from \(120\) units per hour to zero. It then remains stopped for \(1\) hour. Find the total output over the \(7\)-hour period.
For \(2\le t\le 6\), the line's rate is \(120-30(t-2)\) units per hour. This gives \(120\) at \(t=2\), because \(120-30(0)=120\), and gives zero at \(t=6\), because \(120-30(4)=0\). The full model is
Calculate each contribution separately. The initial constant-rate interval produces
For the slowdown interval, an antiderivative of \(120-30(t-2)\) is \(120t-15(t-2)^2\). Thus,
The stopped interval contributes \(\int_6^7 0\,dt=0\). Therefore,
The total output is \(480\) units. To check the slowdown calculation, differentiate \(120t-15(t-2)^2\): its derivative is \(120-30(t-2)\), the stated rate. The slowdown's \(240\) units are also consistent with a linearly declining rate over \(4\) hours: the average of the endpoint rates \(120\) and \(0\) is \(60\) units per hour, and \(60(4)=240\).
Common Pitfall: Using One Rate for the Whole Interval
A rate that changes with time cannot usually be replaced by its endpoint value and multiplied by the entire duration. That calculation assumes the endpoint rate applied throughout, which is a different model.
Worked Example: Correcting an Endpoint-Rate Calculation
Suppose \(r(t)=50+10t\) units per hour for \(0\le t\le 4\). A tempting but incorrect calculation uses the rate at the end, \(r(4)=50+10(4)=90\), and concludes that production is \(90(4)=360\) units.
This is wrong because the rate was not \(90\) units per hour for all \(4\) hours. The correct total is
The correct answer is \(280\) units, not \(360\). Check by differentiating \(50t+5t^2\): the derivative is \(50+10t\), as required. The endpoint-rate calculation overestimates the output because the rate is below \(90\) during the earlier part of the interval.
Choosing and Checking the Model
Before evaluating an integral, identify what the rate describes and which times it covers. The units are a useful guide: if \(t\) is in hours and \(r(t)\) is in units per hour, the integral should have units of units. If a stated rate is instead in units per minute, use time in minutes or convert the rate before integrating; mixing hours and minutes changes the numerical total.
For a production schedule, check that the rate agrees with the description on every interval. At a slowdown, its values should reflect the decline; at a complete stop, it should be zero. Then evaluate each interval's contribution and add them. A final check can differentiate the antiderivative on each continuous piece or compare a constant-rate interval with rate multiplied by duration.
The key idea is simple but powerful: production accumulates at the rate \(r(t)\), so total production over an interval is the integral of that rate. Constant schedules, changing schedules, and stops all fit the same framework once the rate and time intervals are modeled correctly.
Check Your Understanding
For each question, keep track of the time interval and the production-rate units.
- A machine produces at \(75\) units per hour for \(3\) hours. Write an integral for the total and evaluate it.
- If a production rate is \(r(t)=40+5t\) units per hour for \(0\le t\le 4\), what is the total output?
- A line produces at \(100\) units per hour for \(2\) hours, then stops for \(30\) minutes. How many units does it produce over the full elapsed time?
- Why must a linearly decreasing rate be integrated over its slowdown interval rather than replaced by its initial rate?
- A rate model is piecewise defined on three consecutive intervals, with rate zero on the last. How should the total production be calculated?
