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I've always heard physics teachers say, "We'll ignore the integration constant here" whenever they're deriving any formula or solving any question for that matter which involves integration. Just for an example, we say -

Work done a light spring = elastic potential energy gained

Mathematically, $W = \int_{0}^y kxdx$

$\implies W = {1\over 2}ky^2 + c$

Let's just say, the value of $c=25$. Then won't the work done be $25 +$PE (elastic PE)? And that in essence means that the formula $W = {1\over 2}ky^2$ (that most of us are taught) is incomplete right? I just want to know why do we just blatantly ignore that integration constant.

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    $\begingroup$ The integral you put as an example is a "definite" integral, the constant of integration has been taken into account. $\endgroup$
    – user65081
    Aug 6, 2021 at 19:53
  • $\begingroup$ It should say$$\int_0^xkx^\prime\mathrm{d}x^\prime.$$ $\endgroup$
    – J.G.
    Aug 6, 2021 at 20:17
  • $\begingroup$ @J.G. can you elaborate a bit more? I'm asking because I've never seen it written like that. $\endgroup$
    – HarshDarji
    Aug 6, 2021 at 20:35
  • $\begingroup$ @HarshDarji See my answer. $\endgroup$
    – J.G.
    Aug 6, 2021 at 20:51

6 Answers 6

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You only have an integration constant when you do an indefinite integral, an integral which does not come with limits. Hence, the integration constant plays the role of an unknown limit. You ignore integration constants when you do a definite integral.

$W = \int_{0}^x kxdx$ is evaluated at $x$ and $0$, there is no integration constant. The area under the curve $kx$ between $0$ and $x$ is a single quantity with no flexibility. When you have known limits, there is no integration constant.

In this example, you can evaluate the antiderivative $\frac{1}{2}k x^{2} + C$ at x and 0 and subtract, finding that the C cancels out, and moreover the value at $x=0$ is $0$, leaving you with $\frac{1}{2}k x^{2}$.

In general, if $f(x)+C = \int f'(x)dx$ then $\int_{a}^{b} f'(x)dx = (f(b)+C) - (f(a)+C) = f(b)-f(a)$, ignoring the integration constant because it cancels itself out.

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  • $\begingroup$ "You ignore an integration constant when you do an indefinite integral which does not come with limits." Don't indefinite integrals already come without limits? $\endgroup$
    – HarshDarji
    Aug 6, 2021 at 20:12
  • $\begingroup$ Did you mean to say - "You ignore an integration constant when you do an indefinite integral which comes with limits." ? That is, we ignore integration constant when we are doing a definite integral. Right ? $\endgroup$
    – HarshDarji
    Aug 6, 2021 at 20:13
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    $\begingroup$ Thanks, I clarified. Those were important mistypes to catch :) $\endgroup$
    – Alwin
    Aug 6, 2021 at 20:17
  • $\begingroup$ Thanks to you too...you helped me a lot. Cheers! $\endgroup$
    – HarshDarji
    Aug 6, 2021 at 20:35
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At @HarshDarji's suggestion, I'm elaborating on a comment of mine. The fundamental theorem of calculus gives some fairly weak conditions that ensure the following: if $f(a)=b$ and $f^\prime(x)=g(x)$,$$f(x)=b+\int_a^xg(y)dy.$$As others have already noted, physicists aren't guilty of anything here (as long as they get $f(a)$ right), because they're dealing in definite integrals.

But note that I've made the integration variable $y$, or in my comment $x^\prime$, rather than $x$. We mustn't use the same symbol for the integral's upper limit, which is just the free variable $x$ passed to $f$, as we do for the bound integration variable. This is analogous to the fact that $\frac12n(n+1)$ is $\sum_{k=1}^nk$, not $\sum_{n=1}^nn$ (which makes no sense).

Writing $x^\prime$ for it is an especially convenient way to address this concern without forgetting what the integral does.

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  • $\begingroup$ Ohh okay. Then it was my bad. I should've used $x$ instead of $X$ or some other variable. Thank you for clarifying. $\endgroup$
    – HarshDarji
    Aug 7, 2021 at 6:49
  • $\begingroup$ In my view writing $x$ instead of (e.g.) $x'$ is more akin to writing the sum $\sum_{n = 1}^n n$, which---as in the integral situation---doesn't even make sense (unless one adopts additional scope-based rules for the interpretation of such an expression, like one has in a programming language). $\endgroup$ Aug 7, 2021 at 14:06
  • $\begingroup$ @KeeleyHoek Thanks; fixed. $\endgroup$
    – J.G.
    Aug 7, 2021 at 14:24
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Your integral shouldn't have an integration constant because it is definite.

However, what you may be referring to is in the context of potential energy. The absolute value of potential energy is useless. It is arbitrary up to a constant shift. What really matters is the change in that energy. This reflects the work done by a conservative force. For example, if I have a potential energy $U$, and I add a constant to it, $U+c$, then yes this changes the absolute value of $U$. But when I calculate the work done:

$W = - \Delta U = -(U_2 - U_1) = - ((U_2 + c) - (U_1 + c))$

it doesn't matter if I have the constant in there or not. The work done is still the same.

Also, the derived conservative force is the same as well:

${\bf F} = - \nabla U$.

The nabla operator does nothing with the constant, it returns the zero vector when operating on a scalar constant.

Practically, we are interested in work done along a path, or the conservative force. We are never interested in the absolute value of the potential energy, so we "ignore the constant."

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Mathematically, $W = \int_{0}^x kxdx$

$\implies W = {1\over 2}kx^2 + c - ({1\over2}k(0)^2+c)={1\over2}kx^2$

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Let's take a closer look at the specific case that you use as an example: performing integration to arrive at an expression for potential energy.

The thing with potential energy: potential energy doesn't have an intrinsic zero point. In any calculation that you do you are evaluating a difference in potential. In that sense the choice of zero point is arbitrary.

In the case of elastic potential energy of a spring there is a natural choice of point of zero potential: the relaxed state of the spring. But still, no self-contradiction will arise if you assign as potential a non-zero value 'c' to the relaxed state of the spring. Then when you assign a numerical value (for elastic potential energy) to a particular shortened/elongated state of that spring you add that same value 'c' that you assigned to the relaxed state of the spring.

In any calculation: what you use is the difference in potential energy between two states of the spring, so that factor 'c' will always drop out of the calculation.


Here is a case where the choice of zero point is a bit tricky: gravitational potential energy.

One may be tempted to do the following: take two objects, and call the state where the two objects have merged the state of zero potential gravitational energy. Then you can assign a potential to every state where the two objects are at some distance to each other.

The problem with that: if you treat the two gravitating bodies as point masses then they can come arbitrarily close to each other. Gravity is an inverse square law; when the distance becomes infinitely small the force becomes infinitely large. So: mathematically the integration fails: you cannot integrate all the way to distance zero.

The above infinity problem is evaded in the following way: in the case of an inverse square force such as gravity the value of zero potential is put at infinitely far away from each other.

As two objects are gravitationally attracted to each other they accelerate towards each other. They are moving down the gravitational potential, and gravitational potential energy is transformed to kinetic energy.

Between any two distances what is relevant to you is the difference in gravitational potential energy. All you ever use is difference in gravitational potential energy between distance A and distance B. That is why you have freedom to choose where you put the point of zero potential energy.

As two objects accelerate towards each other, accelerated by gravity, the value of the potential energy you assign is a negative value. Can potential energy be negative? Again: that negative value doesn't have an intrinsic meaning. The difference in gravitational potential between some point A and some at larger distance point B is a positive value.

All you ever use is difference of potential energy between two states.


All cases where integration is done, and the integration constant is ignored, are cases where (like potential energy) you are using a difference; a difference between two states. So any integration constant that you add will drop out of the calculation anyway.

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  • $\begingroup$ Wait isnt the PE of a bob 0 at its lowest point? So zero point is the lowest point. Right? I'm so sorry if I'm going on a tangent. But this is literally what I was taught $\endgroup$
    – HarshDarji
    Aug 7, 2021 at 6:11
  • $\begingroup$ @HarshDarji To assign the value of zero potential to the lowest point of a pendulum swing is the natural choice of zero point. It's so natural that it is always so chosen. When people do something in the same way every time they stop thinking about it (which is not good). If you assign a value of 'c' to the potential at the lowest point of the swing, and the highest point of the swing is 'h' higher than the lowest point, then the potential at the highest point is 'c + h'. For the outcome of your calculation only 'h' is important. 'c' is important for bookkeeping, but not for the outcome. $\endgroup$
    – Cleonis
    Aug 7, 2021 at 6:36
  • $\begingroup$ Ahh...okay. So that zero potential thing is more like a convention. Thanks for helping me! $\endgroup$
    – HarshDarji
    Aug 7, 2021 at 6:38
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Sorry for my poor english. My native language is french.

I think you are confusing the notion of Riemann integral and the notion of antiderivative. When we define them, the two have nothing to do with each other.

The Rieman integral is defined as the limit of Riemann sums: we divide the interval into infinitely small elements and we sum the whole passing to the limit. This is what we do when we calculate the potential energy of a spring.

The antiderivative is simply the reverse operation of the derivative. His definition involves an indeterminate constant.

But there is a fundamental theorem of analysis which says that the Riemann integral is the variation of the antiderivative between the bounds.

This theorem is obviously very useful! But we do without it when we calculate an integral numerically. And we could also estimate the area under a curve by weighing the sheet of paper!

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  • $\begingroup$ Thank you. And your english is pretty good :)) $\endgroup$
    – HarshDarji
    Aug 7, 2021 at 10:23

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