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What is a possible visual meaning of the integral of a real function $x(t)$ other than "the area under the graph"?

I'm asking this so that i can avoid thinking about a graph when thinking about an integral, and view the integral as a property of a point in infinite-dimensional space.

In the discrete case of a real sequence $x(n)$ there is the sum, finite or infinite: $x(1) + x(2) + \ldots = x(1)*(1-0) + x(2)*(2-1) + \ldots$

So also, what is the visual meaning of this sum, which, when generalized to a continuous variable, we get the integral.

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    $\begingroup$ Total change. That's it. $\endgroup$ – Sean Roberson Sep 25 '18 at 18:52
  • $\begingroup$ Actually the Riemann integral is not the area under the graph. Not in every circumstance, at least. Areas should be non-negative while functions might be negative. And even by requiring that $f(x)\geq 0$ for any $x\in[0,1]$, that does not grant that $f$ is Riemann-integrable over $[0,1]$. $\endgroup$ – Jack D'Aurizio Sep 25 '18 at 19:07
  • $\begingroup$ @SeanRoberson: do you expect this to be enlightening for the OP ? $\endgroup$ – Yves Daoust Sep 25 '18 at 19:25
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$f(b)=f(a)+\int_{a}^{b}f^{'}(t)dt$

In other words "a later value of the function as the sum of all small changes from another earlier value!"

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Accumulate acceleration to get speed.
Accumulate speed to get distance.
Accumulate width to get area.
Accumulate area to get volume.

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In intuitive terms, an integral can be understood as the average of a function over an interval, or more generally over a domain, times the extent of that domain.

E.g. the area under a curve is the width of the interval times the average "height" of the function.

The gravity center of a 3D shape is the average position of the points, i.e. the integral of the position vector over the volume.

The displacement of a vehicle over a time interval is the average speed times the delay, i.e. the integral of the speed.

An integral accumulates instantaneous variations to yield a global variation.

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Whenever $f$ is the density of some substance, the integral of $f$ is the total amount. This interpretation works well for multiple integrals too.

For example, if $f(x)$ is the electric charge density (charge per unit of length) in a rod $a \le x \le b$, then $\int_a^b f(x) \, dx$ is the total amount of electric charge in the rod.

And if $f(x,y)$ is the electric charge density (charge per unit of area) in a plate of shape $D$ (a domain in $\mathbf{R}^2$), then $\iint_D f(x,y) \, dxdy$ is the total amount of electric charge in the plate.

And if $f(x,y,z)$ is the electric charge density (charge per unit of volume) in a solid of shape $E$ (a domain in $\mathbf{R}^3$), then $\iiint_E f(x,y,z) \, dxdydz$ is the total amount of electric charge in the solid.

Electric charge is nice since it can be both positive and negative. If $f$ is positive, you can (for example) think of mass density instead.

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  • $\begingroup$ I wonder whether physical explanations are the only possible. If f is a finite or infinite sequence or a real function, then the totality of its values define a point in a finite or infinite dimensional space. So what motivates us to pass from this totality of values to the integral? $\endgroup$ – exp8j Sep 26 '18 at 8:51
  • $\begingroup$ @J.Avaris: The integral is used whenever you want to “add up” all those values. Of course you can't actually add uncountably many finite values, so you multiply each value by an infinitesimally small “length element” or “area element” or “volume element” or whatever is suitable, and then you can “add up” (= integrate) all the infinitely many infinitesimal contributions to get a finite value. This reasoning isn't rigorous, but it works well for the purpose of intuition. $\endgroup$ – Hans Lundmark Sep 26 '18 at 9:26
  • $\begingroup$ I think that exactly because we are bound to multiply by something "infinitesimally small" (otherwise the sum would be infinite) that the connection with simply "adding up" of countably many values, is broken. It seems like an artifical adjustment. One could argue that in the countable case, we are implicitly multiplying by 1, or that we are calculating the inner product of the point(or vector) f by the vector (1,1,...). But what would be the meaning of this? $\endgroup$ – exp8j Sep 26 '18 at 10:02
  • $\begingroup$ If you want learn how to view discrete sums as a special case of integrals, look up the Stieltjes integral or, alternatively, the Lebesgue integral with respect to a discrete measure. $\endgroup$ – Hans Lundmark Sep 26 '18 at 10:54

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