The question

From practical experience, I know that the unit of an integral - resulting from integration of an expression with respect to a variable with a unit (i.e. non-dimensionless variable) - is the same as if the expression and the variable were multiplied.

Similarly, the unit of a derivative - resulting from differentiation of an expression with respect to a variable with a unit (i.e. non-dimensionless variable) - is the same as if the expression was divided by the variable.

This makes sense to me, and I have never really thought about it before. The question is: can this fact be regarded as a general (mathematical) rule? If yes; is it possible to give a compelling, rigorous and yet simple argument for this fact? If not; are there any good examples to illustrate why this may not be a general fact?

An example

When calculating the distance travelled as an integral of velocity with respect to time, the units are the same as when multiplying the two quantities.

$$ \int v \, \mathrm{d}t = s $$ $$ [\mathrm{m}/\mathrm{s}][\mathrm{s}] = [\mathrm{m}]$$

In thermodynamics, when calculating the entropy as a partial derivative of the internal energy with respect to temperature, the units are the same as when dividing the former quantity by the latter.

$$ \left(\frac{\partial{U}}{\partial{T}}\right)_{V,\mathbf{n}} = S $$ $$ \frac{[\mathrm{J}]}{[\mathrm{K}]} = \frac{[\mathrm{J}]}{[\mathrm{K}]}$$


This is probably a rather trivial question, and a strongly suspect that the answer is yes based on the relationship between the operations. However, I would love to get a proper answer based on sound mathematical arguments. Any helpful suggestions or hints are greatly appreciated!


One interpretation of the derivative is as the rate of change of a function. If we have a function $f(x)$ taking values with unit $A$ and outputting values with unit $B$, then the derivative $\frac {df}{dx}$ is the rate of change of $f(x)$ (in units of $A$) with regard to the rate of change of $x$ (in units of $B$), that is $\frac {df}{dx}=\frac {\Delta f}{\Delta x}$ as $\Delta x$ becomes infinitesimal. Since $\Delta f$ is in units of $A$ and $\Delta x$ is in units of $B$, $\frac {df}{dx}$ is in units of $\frac AB$.

With respect to integration, we could simply say that as integration is the inverse of differentiation, the unit operation must be the inverse of that of differentiation, i.e. the inverse of division, multiplication. To be slightly more rigorous, consider integration as the sum of rectangles $$\int_a^b f(x)\; dx=\sum_i hf(x_i)$$ approximating the area under the graph. The unit for the integral must be the same as the unit for each rectangle. The area of the $i$th rectangle is $h*f(x_i)$, where $h$ is the width of each rectangle. Noting that $h$ is measured in $A$ and $f(x_i)$ is measured in $B$, we see that the area of each rectangle, and so the entire integral, is measured in $AB$.

  • $\begingroup$ What if the the function is dimensionless? $\endgroup$ – user599310 Jan 24 '20 at 23:37
  • $\begingroup$ @user599310 Then... this is not the case we're talking about? The question was about the case where both the function and the variable over which we are integrating have units. Wrt your comment on Regret's answer, $f(x)=x$ has the same units as $x$, and so if $x$ has units of meters then indeed its antiderivative would have units of meters squared, a unit of area. (Of course, if $x$ has units of, say, mass, then the antiderivative would have units of mass squared, not area.) $\endgroup$ – Laertes Jan 26 '20 at 1:01

I'm no expert in dimensional analysis, but let's take a look at the definition of the derivative:

$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

You can clearly see the numerator is in the units of the return value of the function and the denominator is in the units of the argument to the function. If your function returns a value of unit $\rm A$ and takes an argument of unit $\rm B$, then the derivative will have a return value of unit $\rm A/B$.

  • $\begingroup$ What if the the function has no units? For example, $f(x)=x$ (no units). Then the anti-derivative $F(x)=x^2$ must have dimensions of area? $\endgroup$ – user599310 Jan 24 '20 at 23:52

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