Why an Integral Has Units
A definite integral combines values of a function with small changes in its input. Those changes have units too, so the integral generally has units that differ from the integrand.
For example, velocity is measured in metres per second, while time is measured in seconds. Integrating velocity with respect to time therefore gives units of metres. Geometrically, if a graph’s horizontal and vertical coordinates are both measured in metres, the area under the graph has units of square metres.
These conclusions come from the same rule: multiply the units of the integrand by the units of the variable of integration. The rule is useful both for interpreting a result and for detecting a calculation that cannot be correct.
The Unit Rule
Here \(U\) and \(V\) name units, such as metres per second and seconds. In dimensional notation, we can write \([f]\) for the dimensions or units of \(f\), and \([x]\) for those of \(x\). The rule is then expressed as
The product is essential. The integral does not, in general, have the same units as \(f(x)\), nor does it have the units of \(x\) alone. If the input is dimensionless, however, multiplying by its units does not change the units of the integrand.
Why the Units Multiply
For a Riemann-integrable function \(f\) on \([a,b]\), a Riemann sum has the form
where each \(\Delta x_i\) is the width of a subinterval and \(x_i^*\) is a sample point in that subinterval. Each term is a function value multiplied by an input width. Its units are therefore \([f]\,[x]\). Adding terms with the same units preserves those units, and the definite integral is the limit of these sums.
This also explains why the differential \(dx\) contributes the units of \(x\): it represents an input width in the Riemann sums. The notation \(f(x)\,dx\) thus signals a product of the integrand’s units and the input’s units.
Worked Example: Integrating Velocity Over Time
Worked Example: Integrating Velocity Over Time
A constant velocity is \(3\ \text{m/s}\) over the time interval from \(0\ \text{s}\) to \(5\ \text{s}\). Find the definite integral of velocity with respect to time and state its units.
Let \(v(t)=3\ \text{m/s}\). The time interval has length
The function is constant, so the integral is the constant value multiplied by the interval length:
The unit check is \((\text{m/s})(\text{s})=\text{m}\), because the seconds cancel. The result has units of length, as the unit rule predicts: velocity units multiplied by time units give metres.
As a check on the numerical calculation, the graph of this constant function is a rectangle with height \(3\ \text{m/s}\) and width \(5\ \text{s}\). Its signed area is \(3\cdot5=15\), and its area units are \((\text{m/s})(\text{s})=\text{m}\), matching the integral.
Worked Example: A Flow Rate That Changes with Time
Worked Example: A Flow Rate That Changes with Time
Suppose a flow rate over \(0\leq t\leq4\) minutes is described by
where \(t\) is measured in minutes. Find the integral of this rate over the four-minute interval, including its units.
First check that both terms in \(q(t)\) have units of litres per minute. In the second term, \((\text{L/min}^2)(\text{min})=\text{L/min}\). The integrand is therefore consistently a flow rate.
For the numerical integration, use the time measured in minutes and integrate the corresponding numerical expression from \(0\) to \(4\):
The rate is measured in litres per minute and the time variable in minutes. Hence the integral has units \((\text{L/min})(\text{min})=\text{L}\). The result is \(12\ \text{L}\).
To check the antiderivative calculation, differentiate \(2t+0.25t^2\):
This recovers the numerical expression for the rate. The units of the original rate and time multiply to litres, which confirms the units of the integral.
Worked Example: A Geometric Area
Worked Example: A Geometric Area
Let \(x\) be horizontal distance in metres, and let a straight-line graph have height \(h(x)=0.5x\) metres for \(0\leq x\leq4\ \text{m}\), where \(x\) in the formula is the numerical distance in metres. Find the area represented by the integral and state its units.
Since the input is a length and the graph height is also a length, the integral units are \(\text{m}\cdot\text{m}=\text{m}^2\). Numerically,
Thus the area is \(4\ \text{m}^2\). To check geometrically, the graph rises from \(0\ \text{m}\) to \(0.5(4)=2\ \text{m}\), forming a triangle with base \(4\ \text{m}\) and height \(2\ \text{m}\). Its area is
The geometric calculation agrees with the integral, including both the numerical value and the square-metre units.
A Proof of the Unit Rule
The Riemann-sum definition justifies the unit rule directly. Suppose each function value \(f(x_i^*)\) has units \(U\), and each subinterval width \(\Delta x_i\) has units \(V\). Then every product in the sum has units \(U\cdot V\). Because all the terms share these units, their sum also has units \(U\cdot V\).
When the mesh of the partition shrinks and the Riemann sums converge to the definite integral, the limit retains the units of the quantities being summed: the numerical values approach a limit, while the measurement represented by each sum is still in units \(U\cdot V\). Therefore the integral has units \(U\cdot V\). This argument presumes that \(f\) is Riemann integrable on the interval, so that the definite integral exists.
Using Units to Check a Calculation
Units provide a quick consistency test, but they do not by themselves prove that a numerical answer is correct. A result can have the right units and still contain an arithmetic error. Conversely, if its units are wrong, the calculation or the stated interpretation needs attention.
For example, the velocity calculation above should not produce a result in metres per second. Multiplying \(3\ \text{m/s}\) by \(5\ \text{s}\) cancels seconds, so the result must be in metres. A claim of \(15\ \text{m/s}\) would fail this unit check even though the numerical multiplication \(3\cdot5=15\) is correct.
Another frequent source of confusion is mixing units for the input. If a time interval is given in minutes while a rate is in litres per second, convert one of them before combining the numerical values. For instance, \(2\) minutes is \(120\) seconds; multiplying a rate in \(\text{L/s}\) by \(2\) as though the time were in seconds would give an incorrect result.
Units and Geometric Area
The geometric area interpretation gives a useful way to remember the rule. The horizontal width has the units of the input, and the vertical height has the units of the function. A rectangle formed from those measurements has area units equal to their product. Riemann sums add many such products, so the units of the integral follow the same pattern.
This applies even when the units are not both lengths. A graph of velocity against time has rectangle units \((\text{length}/\text{time})(\text{time})\), which reduce to length. A graph of height against horizontal distance has rectangle units \((\text{length})(\text{length})\), which reduce to area. The visual area under a graph is therefore measured using the units of both axes, not simply by the units printed on the vertical axis.
If the graph passes below the horizontal axis, the signed integral can be negative, as covered in the earlier material on signed area. Its units are still the product of the vertical and horizontal units: a minus sign changes the sign of a quantity, not its units.
Summary
To determine the units of a definite integral, identify the units of the function values and the units of the input. Multiply them. The Riemann-sum terms \(f(x_i^*)\Delta x_i\) show why: each term combines a function value with an input width. Check that the bounds use the input’s units, and convert measurements before carrying out numerical calculations.
Check Your Understanding
For each question, state the units carefully and explain the unit multiplication where appropriate.
- A constant acceleration is measured in \(\text{m/s}^2\). What are the units of its definite integral with respect to time \(t\), measured in seconds?
- A function gives a height in centimetres as a function of horizontal distance in centimetres. What units does its definite integral have?
- A flow rate is measured in \(\text{L/min}\), and the input variable is measured in minutes. What units does the integral have?
- Someone claims that integrating a velocity measured in \(\text{m/s}\) over a time interval measured in seconds gives a result in \(\text{m/s}\). Identify and correct the error.
- Why do the terms in a Riemann sum provide a justification for the unit rule?
