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Two questions that are greatly lingering on my mind:

1.

Integral is all about area(as written in Wolfram).

But what about indefinite integral? What is the integral about it?? Is it measuring area?? Nope. It is the collection of functions the derivative of which give the original function and not measuring area. So, why "integral"?? And what about the indefinite??? It is not measuring an infinite area ; just telling about the original functions. So, what is the logic of this name??

  1. Famous statement:

    Differentiation breaks apart the function infinitesimally to calculate the instantaneous rate of change, while, on the other hand, integration sums up or integrates the infinitesimal changes to measure the whole change or area .

Yes, totally correct but in case of definite integrals,where small changes are summed up to give the area. But how is the statement related with indefinite integral?? Do they sum up small changes??? What is the connection between them???

I am confused. Please help me explaining these two problems.

  • A definite integral has a definite area to calculate. For an indefinite integral, the right and left bound for the area are not decided. – Arthur Nov 19 '14 at 15:27
  • The connection between them is given by the fundamental theorem of calculus. – Dustan Levenstein Nov 19 '14 at 15:42
  • @Arthur: Indefinite Integrals do not measure the areas;they are functions. –  Nov 19 '14 at 15:43
  • @user36790 If $F(x)$ is the indefinite integral of $f(x)$, then $g(a, b) = F(b)-F(a)$ is a function that gives you the area once the right and left boundaries of the region has been decided. – Arthur Nov 19 '14 at 15:53
  • @arthur, when you are talking about areas, it is always definite integral even when the bounds are not decided (for example an F(x) that is defined by the area of function 'f' from minus infinity till x) – avner Jan 12 '18 at 07:10
  • @avner I can't believe you bumped a three year old question for that one comment. Damage done, I guess. I think of indefinite integrals as really definite integrals with unknown bounds. So when I see $\int f(x)dx$ I think $\int_a^x f(t)dt$. (At least, that's what I think whenever I'm bothering to think about the theory; if it's just a simple integration problem I do think about it as an antiderivative operation). Note that this includes non-existent $a$ so as to not limit the constant of integration needlessly. I don't remember, but I believe that that is what I meant with my previous comment. – Arthur Jan 12 '18 at 07:35
  • @arthur - I found this discussion after I was bothered by this question. I wasn't happy with the answer here because I started to "learn" the subject and reached some level of understanding. At that phase I replied to you. Later I reached deep understanding and then I wrote my understanding as a second answer - see it below :-) Cheers, Avner – avner Jan 14 '18 at 19:11

2 Answers2

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A primitive of a function $f$ is another function $F$ such that $F'=f$. If $F$ is a primitive of $f$, so is $F+C$ for any constant $C$, the so called constant of integration. The indefinite integral of $f$ can be thought of as the set of all primitives of $f$: $$ \int f=F+C. $$ Why indefinite? Because is there some indefinition due to the constant $C$.

What is the relation to areas, or definite integrals? The fundamental theorem of calculus. If $F$ is a primitive of $f$ then $$ \int_a^bf=F(b)-F(a). $$

Indefinite integrals are a tool for the computation of definite integrals.

Julián Aguirre
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Assuming you have a function f that is integrabile in a segment, then for any value of 'a', you can define a definite integral function by: $$ F(x) = \int_a^x f $$ that is defined by the area under f in the range [a,x]

Hence, we can say that all of them together (for all possible values of 'a') form indefinite integral of f - marked by $$ \int f(x)dx $$ Assuming f is continues in the segment, we can show - base on the derivative definition - that each of these definite integral functions is a primitive function of f (this is part of the fundamental theorem of calculus).

Next, it is easy to show that all primitive functions of f are of the form F(x) + C, where C is a constant and F'(x) = f(x).

Hence, we define the indefinite integral of f as all primitive functions of f, and we mark $$ \int f(x)dx = F(x) + C $$

avner
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