# Integral as area under a curve

Let $$f(x)=x.$$ If the units of $$x$$ are meters, then the area under the curve $$y=f(x)$$ between $$x=1$$m and $$x=2$$m is $$\frac12(2\text m)^2-\frac12(1\text m)^2 = 1.5\text m^2.$$

Now let $$f(x)=x^2.$$ Again, the units of $$x$$ are meters. This time, the area under the curve $$y=f(x)$$ between $$x=1$$m and $$x=2$$m is $$\frac{(2\text m)^3}{3} - \frac{(1\text m)^3}{3} =3.5\text m^3.$$ The area gets expressed in cubic meters. Where is the mistake?

What happens with the units when the function is $$f(x) =e^x$$ or $$f(x)=\frac{1}{x}$$?

When you think of the integral $$\int_a^b f(x)dx$$ as the area under the graph of $$f$$ over the interval $$[a,b]$$, the variable $$x$$ has units Length. So does the dependent variable $$y$$, whose value at $$x$$ is $$f(x)$$: the height of the graph at that point. That is independent of the unit calculations suggested by a formula for $$f$$.

Since $$dx$$ (which represents a change in $$x$$) also has units Length, the integrand $$f(x)dx$$ has the units $$\text{Length}^2$$ as expected.

In the example $$\int_a^b x^2dx = \frac{b^3}{3} - \frac{a^3}{3}$$ you can think of $$\frac{b^3}{3}$$ as $$\frac{b^2}{3} \times b$$ which will have units $$\text{Length}^2$$ since the numerator of the fraction has units Length.

In an application where $$f(x)$$ is the velocity at time $$x$$ the integrand has units $$\frac{\text{Length}}{\text{Time}}\times \text{Time} = \text{Length}$$ as expected for the distance covered during the time interval.

• @ryang That's another approach. You still have to note that the differential $dx$ has units $m$. May 28 at 15:38
• I do not see why $f(x)$ needs to have the same units as $x$. If $x$ is a distance measured in $m$ and $f(x)$ a force measured in $N$, then the integral area can be work (i.e. energy) measured in $Nm$ (i.e. $J$) May 29 at 1:04
• @Henry When thinking of the integral as the area, both $x$ and $f(x)$ have units Length. The answer explicitly offers another application. May 29 at 1:14
• @Henry $f(x)$ does not need to have the same units as $x$, unless you want to interpret the integral as an area. In which case, it does. May 29 at 2:15
• Your’re absolutely right. My (now-deleted) remark was regarding the OP’s underlying question rather than their actual question. May 29 at 6:40

Area = $$\int y\cdot dx;$$ So each of $$(x,y)$$ has a physical dimension/degree one, the LHS and RHS for Area and integral both have dimension two.

Accordingly we should always have $$y=y(x)$$ with dimension unity. Else the physics dimension does not tally.

Examples of $$y(x)$$ having permissible unity dimension are: $$a, x^2/b, \frac{x^2-a^2}{x+d},x+\frac{p^2}{x}, x e^{x/q}, \frac{p^2}{x}, x \log \lambda$$

where the variable $$x$$ symbols $$a,b,p,q$$ represent a constant of unity dimension... except $$\lambda$$ which is dimensionless.

Differential of any variable has dimension of that variable itself.

For example differentiating w.r.t $$x$$ reduces the degree of $$y$$ by one.

$$y(x)= f(x)= \frac{x}{\lambda} +\frac{p^2}{x}$$

$$\frac{dy}{dx}= f'(x)=\frac{1}{\lambda}-\frac{p^2}{x^2}$$

In physics the variables are added on but dimensions must tally: By Newton's Law, Force is mass times rate of velocity or mass rate times the velocity.

$$M L /T^{-2} \text{ or } (M/T) L /T^{-1}..$$

There are two paradigms when it comes to the treatment of units (note that this is a generalisation as some mathematicians use the physicist's paradigm, and vice versa):

Variables ($$x$$, $$t$$, etc.) are the numerical value of quantities when expressed in a chosen set of units.

• $$x$$ is the number of metres, $$t$$ is the number of seconds, etc.
• $$x = 1$$ represents the physical length of 1 metre, $$t = 1$$ represents the physical length of 1 second, etc.

In this paradigm, there are no units in any of the symbols $$x$$, $$y$$, $$f$$, etc.

So when you write $$y = x^2$$, both sides of the equation don't have units. The variables $$x$$ and $$y$$ are numbers that represent a length, and so the result of the integral is a number that represents an area.

Variables ($$x$$, $$t$$, etc.) are actual physical quantities, and therefore contain units within them.

• $$x$$ is the number of metres multiplied by a physical metre; $$t$$ is the number of seconds multiplied by a physical second, etc.
• $$x = 1 \,\mathrm{m}$$ is a physical length, $$t = 1 \,\mathrm{s}$$ is a physical time, etc.

This paradigm is unit-independent; e.g. $$x = 1 \,\mathrm{m}$$ and $$x = 100 \,\mathrm{cm}$$ mean the same thing.

Under this paradigm, if $$y$$ is supposed to be the height of the curve, then you cannot write $$y = x^2$$, because area is not length. Instead, you must write

$$\frac{y}{\mathrm{m}} = \left( \frac{x}{\mathrm{m}} \right)^2$$

or

$$y = \frac{x^2}{\mathrm{m}}$$

so that the dimensions are consistent.

Likewise:

• $$y = e^x$$ must instead be $$y = e^{x/\mathrm{m}} \,\mathrm{m}$$ (see why the argument of an exponential must be dimensionless).
• $$y = 1/x$$ must instead be $$y = \mathrm{m}^2 / x$$.

On the other hand, if you aren't thinking about the integral as being an actual area, then you can have things like

• $$s = \int v \,\mathrm{d}t$$ (displacement = velocity × time)
• $$W = \int F \,\mathrm{d}s$$ (work = force × displacement)

so long as the dimensions of the equation are consistent (integral = integrand × independent variable).