From a Rate to a Total
A rate describes how quickly a quantity changes as an independent variable changes. It might describe water entering a tank per minute, material deposited per centimetre, or energy used per hour. If the rate varies across an interval, one rate value alone cannot describe the total accumulated over that whole interval.
The definite integral provides the appropriate total: it adds the contributions from all the small parts of the interval. To use it correctly, identify three things before calculating: the rate function, the variable with respect to which the rate is given, and the interval over which accumulation occurs.
Here, the nonnegative-rate assumption matches the ordinary idea of adding accumulated quantity. The variable \(x\) is a placeholder: in an application it might be time, position, distance, or another quantity. The differential \(dx\) indicates that the accumulation is taken with respect to \(x\).
Why the Integral Gives the Total
Consider a short interval of width \(\Delta x\) near a value \(x\). If the rate there is approximately \(r(x)\), then the amount accumulated over that short interval is approximately the rate multiplied by the interval width:
To estimate the total from \(a\) to \(b\), divide the interval into smaller pieces. On each piece, use a representative rate value and multiply it by the width of that piece. Adding these products gives an approximation to the total.
More precisely, take a partition \(a=x_0<x_1<\cdots<x_n=b\), and write \(\Delta x_i=x_i-x_{i-1}\). Choose a sample point \(\xi_i\) in each subinterval \([x_{i-1},x_i]\). The accumulated amount on that subinterval is approximated by \(r(\xi_i)\Delta x_i\), so the entire interval is approximated by the sum:
As the partition becomes finer, each piece is shorter, so the rate varies less across each individual piece. For a continuous rate, these sums approach a limit. That limit is the definite integral, which establishes why the integral represents the accumulated total:
This argument is the Riemann-sum definition of the definite integral applied to a rate. It explains the method: multiply rate by a small interval width to estimate a small contribution, add the contributions, and take the limit as the pieces become arbitrarily small.
A Reliable Setup
Before evaluating an integral, translate the situation into its mathematical parts. Keep the bounds tied to the interval in the problem, and use the variable with respect to which the rate is defined.
Determine what total the question asks for, such as a volume of water, a mass, or an amount of energy.
Record the function that gives the rate and the independent variable used in that function.
Use the beginning and ending values of the interval over which the quantity accumulates.
Integrate the rate with respect to its variable across the stated interval, then describe the result as a total accumulated quantity.
The integral is not just a calculation to perform after reading a word problem. It is the mathematical statement of which rate is being accumulated, over what interval, and with respect to which variable.
Worked Applications
Worked Example: Water Flowing into a Tank
Water enters a tank at a rate \(r(t)=4+3t\) litres per minute, where \(t\) is the number of minutes after the observation begins. Find the amount entering during the first \(2\) minutes.
The rate is given as a function of time, and the interval is \(0\le t\le 2\). Therefore the accumulated amount is represented by:
An antiderivative of \(4+3t\) is \(4t+\frac{3}{2}t^2\), since differentiating gives \(4+3t\). Evaluate at the bounds:
Thus \(14\) litres enter the tank during the first \(2\) minutes. To check the integration, differentiate the antiderivative: \(\frac{d}{dt}(4t+\frac{3}{2}t^2)=4+3t\), the given rate. The result is a total over the interval, not the rate at either endpoint.
Worked Example: Material Deposited Along a Strip
A strip receives material at a rate of \(\lambda(x)=2x+1\) grams per centimetre at position \(x\) centimetres, for \(1\le x\le 4\). Find the total mass deposited along this part of the strip.
The rate is specified by position, not by time. The required total is therefore the integral with respect to \(x\), from position \(1\) to position \(4\):
An antiderivative is \(x^2+x\), because its derivative is \(2x+1\). Substituting the upper and lower bounds gives:
The total deposited mass is \(18\) grams. The check is \(\frac{d}{dx}(x^2+x)=2x+1\), which recovers the specified rate. This example also shows why the independent variable need not represent time: the rate can describe accumulation across a spatial interval.
Worked Example: Medicine Delivered by an Infusion Pump
An infusion pump delivers medicine at rate \(q(t)=8+2t+t^2\) milligrams per hour, where \(t\) is measured in hours. Find the amount delivered from \(t=0\) to \(t=3\).
The rate function is \(q(t)\), and the interval is from \(0\) to \(3\) hours. Set up the accumulated amount as:
An antiderivative is \(8t+t^2+\frac{t^3}{3}\), since the derivative of each term is \(8\), \(2t\), and \(t^2\), respectively. Evaluate it at both endpoints:
The pump delivers \(42\) milligrams during the three-hour interval. As a check, differentiating \(8t+t^2+\frac{t^3}{3}\) gives \(8+2t+t^2\), exactly the stated delivery rate.
Common Pitfall: Confusing a Rate with a Total
A rate value describes what is happening at one value of the independent variable. It does not, by itself, give the total across an interval. For example, in the water-flow problem the rate at \(t=2\) is:
A mistaken calculation might report \(10\) litres as the amount that entered during the first \(2\) minutes. That number is the rate at the endpoint, not the accumulated amount. The correct total uses the rate over the entire interval:
The correction is to distinguish the value \(r(2)\) from the integral \(\int_0^2 r(t)\,dt\). The first is an instantaneous rate at one time; the second combines the varying rate across the specified interval.
Reading the Integral as an Application
When setting up an accumulation integral, make sure each part has a role in the situation. The expression being integrated is the rate, the differential records the variable of accumulation, and the bounds specify where accumulation starts and stops. Changing one of these can describe a different total.
These integrals are useful whenever a quantity builds up through a changing process. Examples include a changing flow into a container, material distributed along a length, and medicine delivered at a varying rate. In each case, the same reasoning applies: use the local rate to describe small contributions, add those contributions over the interval, and represent the limiting total with a definite integral.
Check Your Understanding
For each question, identify the rate, variable, interval, and integral where appropriate.
- A stream flows into a pool at rate \(5+t\) litres per minute for \(0\le t\le 4\). Write the integral for the amount entering.
- A material is deposited along a wire at rate \(3x\) grams per centimetre from \(x=2\) to \(x=5\). Which variable should appear in the differential, and what integral represents the total?
- In a rate model \(r(t)\), explain the difference between \(r(3)\) and \(\int_0^3 r(t)\,dt\).
- Explain why dividing an interval into small pieces leads to products of the form \(r(\xi_i)\Delta x_i\).
- A rate function is given in terms of position \(x\), but an integral is written using \(dt\). What should be checked before using that setup?
