Integral as area under a curve

Question

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 $f(x)=\frac{1}{x}$?

Answer 1

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.

Answer 2

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.

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

Answer 3

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):

1. Mathematician's paradigm

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.

2. Physicist's paradigm

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).