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I have been introduced to calculus starting from limits. So from my point of view I have been considering that for $$\int_a^b{f(x) dx}$$ $x$ starts from a very closer value of $a$ such that $x→a$ and stops at $x→b$

But I have also known that at the very beginning of calculus there did not prevail any notion of limits. So from that point of view $x$ should start from $x=a$ to $x=b$.

So which perspective is correct? Does $x$ overlap those limits or $x$ don't?

Context of this question: If a function having limits for which the function itself is undefined then what would be the condition of integral?

It will be really helpful if this question is answered based on the difficulty of a high school student. I mean if a high school can understand the answer without any difficulty. Thanks.

MSKB
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    When you learn about Riemann integrals, you will find that the value of $x$ at the endpoints does not affect the integral. So for a high school student I'd say the function is integrated between $x=a$ and $x=b$. – Landuros Sep 17 '21 at 06:13
  • If I am not wrong, you meant the second perspective, right? – MSKB Sep 17 '21 at 06:16
  • Well if that is the case does $\int_1^0{lnx dx}$ make any sense? – MSKB Sep 17 '21 at 06:22
  • This is like asking "When we compute the area of a rectangle, do we include the perimeter?" The perimeter is has no area since it is one-dimensional, so we don't need to think about it. – Jair Taylor Sep 17 '21 at 07:10
  • But the length(height) of the rectangle is dependent on that parametre if I am not wrong – MSKB Sep 17 '21 at 07:44
  • The integral introduced in a first course of calculus is Riemann integral and then the function $f$ being integrated must be defined on $[a, b] $ (the closed interval on which the function is being integrated) and it must be bounded on $[a, b] $. The function $f(x) =\ln x$ is not bounded in $[0,1]$. – Paramanand Singh Sep 17 '21 at 12:54
  • Moreover if the function is not defined at a finite number of points of the interval then we are at liberty to define the function at those exceptional points in any manner whatsoever without impacting the integral in anyway. Thus $\ln x$ not being defined on $x=0$ is not a problem but it is unbounded on $[0,1]$ and that is a problem. – Paramanand Singh Sep 17 '21 at 12:55
  • Your confusion arises primarily because a first course in calculus usually avoids definition of any concepts like limit or integral and hopes that the student will manage somehow. – Paramanand Singh Sep 17 '21 at 12:58
  • Yes perhaps that could be the reason you mentioned in your last commen @Paramanand Singh.......I have been also facing difficulty regarding this question https://math.stackexchange.com/questions/4246182/problems-regarding-certain-approaches-to-some-certain-integrands – MSKB Sep 17 '21 at 13:04
  • So the interval at which I need to integrate the lnx function is (0,1] considering it would irrational to integrate within the interval [0,1] right? – MSKB Sep 17 '21 at 13:06
  • No. Riemann integral is defined only on closed intervals. The integral in question $\int_0^1 \ln x, dx$ can be managed by using improper Riemann integrals which a bit different thing than a Riemann integral. Also removing troublesome point $0$ does not help here as $\ln x$ is unbounded on $(0,1]$. Remember individual points don't play a role in deciding bounded/unbounded nature of a function. – Paramanand Singh Sep 17 '21 at 13:07
  • I am unfamiliar with both of those integrals..... I mean I never came across those terms in my complete high school course on calculus until recently on this forum – MSKB Sep 17 '21 at 13:09
  • See https://math.stackexchange.com/a/1774974/72031 – Paramanand Singh Sep 17 '21 at 13:11
  • And also https://math.stackexchange.com/a/2982757/72031 – Paramanand Singh Sep 17 '21 at 13:14
  • Does high school calculus include Reimann Integrability by any chance which I might have missed or is it a topic for college students not for high school students? I am bit frustrated since it seems like I am lacking enough knowledge about this stuffs – MSKB Sep 17 '21 at 13:27

2 Answers2

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Integration, chronically, started as the technique for finding areas under curves. Although, the approach was not rigorous and definitely not an $\epsilon-\delta$ type, it won't be correct to say that it was entirely different.

Regarding a good (or correct) approach to understanding and handling integrals, Riemann's integration is the answer. Get hang of the concepts of partitions, meshes or norms, Darboux sums, upper and lower sums, and you are on your way.

F. A. Mala
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You can think of it either way (or both ways). We say that the limits of integration are "$x=a$" and "$x=b$", but this is just terminology; it doesn't mean the value of $\int_a^bf(x)dx$ needs to depend on $f(a)$ or $f(b)$. The integral notation has a precise definition, and we can prove that this definition gives the same result regardless of the function values at $a$ and $b$.

Karl
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  • If f(a) is defined then both the approaches should yield almost the same output but what if f(a)/f(b) is undefined? – MSKB Sep 17 '21 at 07:04
  • You can then calculate the limit $\lim_{c\to a}\int_c^b f(x)dx$, e.g. take a function and calculate the integral to $\infty$. $f$ is not defined at infinity (as it is only defined on $\mathbb{R}$), but you can let the limits of integration tend to infinity. – LegNaiB Sep 17 '21 at 07:08
  • So its like following the first approach I mentioned in the question right? – MSKB Sep 17 '21 at 07:50