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I completely understand why we add C when we want an anti-derivative, but finding anti-derivatives is not integration, integration is meant to caculate the accumulations of values or the area under the curve, so do we have to add C when Integrating? (not anti-differentiating)

Does the integration constant come up when integrating from first principles? If not, why are definite integration defined using anti-derivatives instead of being defined from the "first principles of integration"?

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    Hi, welcome to Math SE. Although any area is expressible as a definite integral, that's not actually the point of definite integrals, let alone indefinite ones. With which definitions of each type of integral are you familiar? Because if you asked me for a first-principles definition of a definite integral, mine would be in terms of limits of families of sums, and then one would prove how this is connected to antiderivatives, areas and so on. – J.G. Dec 28 '23 at 17:28
  • Nobody "adds C". It is a notation and as such a convention. It is meant to express that the equation $y'(x)=f(x)$ has only unique solutions up to a constant summand. However, notations also depend on context. You won't find $\ldots + C$ in a scientific publication. Your question has no meaningful answer as long as its context isn't specified. Considering an integral as a volume element is a rather narrow perspective. – Marius S.L. Dec 28 '23 at 17:35
  • @J.G. Well, yes the area is not the definite integral, but the accumulations IS, the definition that i know (taking into account that i don't know much) is basically a limit to an infinite sum, so i think it's the same thing you're describing – Moaaz Mostafa Dec 28 '23 at 17:40

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I'll try to answer some of your questions, albeit out of order.

(1) "Why are definite integration defined using anti-derivatives?" This is because we have the Fundamental Theorem of Calculus. It is usually broken up into two parts for which in this context we would apply part II: (we have some conditions) and the definite integral can be evaluated as the difference of the antiderivative $F(x)$ at the endpoints. $$\int_a^b f(x)dx = F(b)-F(a).$$

This $F(x)$ may or may not be an elementary function. If you put $+C$ they will simply get cancelled out by the subtraction.

Definite integration is an analytic and geometric problem, whereas indefinite integration is an algebraic problem. The use of the $+C$ in $$\int2xdx = x^2 +C$$ communicates that we have a family of functions which could be the antiderivative---it is indefinite. Because when we differentiate the RHS, we get $2x$. Honestly, the $+C$ is overrated and often made fun of. Some teachers will even mark you wrong if you forget the $+C$ in the previous example, but this nonsense. In Integral Bees, such as the infamous MIT Integration Bee, they don't care about the $+C$ because any antiderivative will do to solve the integral (expressed as an elementary function).

The use of infinite sums to compute areas came before the FTC, and probably before the problem of indefinite integration, so there was no $+C$ around.

Finally, where is the $+C$ important? It is important when we have some type of initial conditions like in differential equations or integral equations.

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    Good observation about needing the $+C$ when satisfying initial (and boundary) conditions when solving differential equations. +1. – JonathanZ Dec 28 '23 at 20:35
  • I would say (somewhat equivalently) that the $+C$ is important when you're going to do something else with the integral besides calculate a definite integral. Such as when you solve a DE like $y' = y$ and you go from $\ln y = x + C$ to $y = A e^x$. – ConMan Dec 29 '23 at 02:27
  • actually that's why i asked the last question, since C is going to cancel out, why do we add it when integrating.(again, integrating not antidifferentiating ) – Moaaz Mostafa Dec 29 '23 at 11:03
  • When you do definite integration, you don't put the $+C$. Otherwise, show us an example that does. – Matthew Albano Dec 29 '23 at 12:58
  • I disagree that it is nonsense that a missing "+C" is marked as an error. This "+C" is important because the antiderivate is not unique. For definite integrals , this is of course irrelevant since the constant cancels out anyway. – Peter Jan 24 '24 at 10:59
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"... why are definite integration defined using anti-derivatives instead of being defined from the "first principles of integration"?

They aren't defined that way, they are computed that way. And they're computed that way because it's so much easier to do it that way.

Here's a brief discussion: Is Integrate antidifferentiate?

And here's a more detailed discussion: Understanding The Fundamental Theorem of Calculus, Part 2

JonathanZ
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    Since when are definite integration defined using anti-derivatives? I know that was what the OP said, but you didn't correct them either and went with it. – Divide1918 Dec 28 '23 at 18:17
  • @Divide1918 - Ooh, good catch, thanks. I'll confess that because the two linked answers are all about the distinction, I quickly grabbed a quote from the question without fully parsing it. I've updated my answer to include your observation. Thx again. – JonathanZ Dec 28 '23 at 20:27
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The indefinite integral is the anti derivative.

The definition that you hav in mind is the definition of the (Riemann) integral. If you look in the link such integral is defined over an interval.

the definition of the indefinite integral of a continuos function $f(t)$ is the function $y(x)$ such that $$ y'(x)= f(x) $$ see e.g. this

Such definition happen to coincide with the Riemann integral (at least in the case $f$ is continuos)

Marco
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I want to add a perspective that does not use volumes to consider integrals.

An integral is a solution to a differential equation. Differential equations can be visualized by their corresponding vector field and the integral can be considered as a flow through this vector field. There are many possible flows through a vector field, all depending on where we start. The constant $C$ can be regarded as the parameter that determines the starting point. predator-prey system

Marius S.L.
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