<p>Hi i want to understand the geometric intuition for why there is no constant of integration with definite integrals.</p>
<p>The Algebraic intuition is fine.</p>
<p><span class="math-container">$\begin{align} \int_0^ax^3\,\mathrm{d}x &amp;=\left(\frac14a^4+C\right)- \left(\frac140^4+C\right)=\frac14a^4\end{align}$</span></p>
<p><strong>Question</strong></p>
<p>If our integral is giving us the area under a curve, then surely the original function <span class="math-container">$f(x)$</span> has a different area under it's curve than some other related function <span class="math-container">$f(x) + C$</span> (Diagram attached at the end). Therefore  even when integrating over a range, we would still need to think about an arbitrary constant C.</p>
<p><strong>Solution:</strong></p>
<p>I think I need to separate the interpretations of an integral. Undoing differentiation (returning us to our original function). Vs giving us the area under a curve.</p>
<ol>
<li><strong>Indefinite Integral:</strong> This is getting us back to the function that was differentiated, which will have some arbitrary constant C somewhere. The C does not represent extra potential area, just the possible lost constant when differentiating.</li>
<li><strong>Definite integral</strong>: Is literally asking for the area under that specific function (The integrand) that is being integrated, within that range of integration.  Hence no arbitrary constant C is needed?</li>
</ol>
<p>Although splitting my understanding of an integral into these alternate but related interpretations doesn't feel quite right. Thoughts?</p>
<p><a href="https://i.sstatic.net/s3Gia.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/s3Gia.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-04-29T10:53:39.930Z
Your observations are very keen, but you are re-discovering the Fundamental Theorem of Integral Calculus: you are in good company, with excellent mathematicians as Newton and Leibnitz. The fundamental Theorem of Integral  Calculus links definite integral with indefinite integral/derivative, and it is really fundamental. It is impossible to answer this question without reporting here some pages of  a mathematical analysis book.
BakerStreet
CC BY-SA 4.0
2023-04-29T10:57:04.650Z
This playlist has some videos on integration which I believe can answer your question: https://youtube.com/playlist?list=PLUJGfL_499TI4ihvXuiI8CFUbpp4BawdP
Amit
CC BY-SA 4.0
2023-04-29T11:02:22.633Z
@ Cormjack To have an idea, only think  that the *definition* of *definite* integral has nothing to do with the formula you wrote at the beginning of your question, and nothing to do with differntiation. The link with differentiation is given subsequently, precisely with the Fundamental Theorem.
BakerStreet
CC BY-SA 4.0
2023-04-29T11:03:22.577Z
Ahh great to see you both as usual, thanks for the comments.
CormJack
CC BY-SA 4.0
2023-04-29T11:04:13.733Z
@ You are welcome!
BakerStreet
CC BY-SA 4.0
2023-04-29T11:05:50.933Z
Hahah yes BakerStreet, I feel like self teaching this stuff, is just permanent rediscovery of things people solved a billion years ago. Your observation is actually very astute, because I remember when I took may advanced calculus class, and I learnt all about integration from first principles and the fundamental theorem of calculus etc. I remember being able to do the mechanics, but not really getting the point of it. I felt at the time it was just telling me things I already new. I.e. undoing differentiation and an area under a curve. But now I think I'm realising some subtleties.
CormJack
CC BY-SA 4.0
2023-04-29T11:06:23.830Z
Like I didn't really get what was so fundamental about it. I think now if I review that material and the playlist Amit has provided I wil appreciate it more.
CormJack
CC BY-SA 4.0
2023-04-29T11:08:01.350Z
Yes your last comment BakerStreet is what I was thinking. I will try review this material soon and if I have time post an answer here. (I have another integration question I'm going to post today as well...but it has more economics in it so is a bit more justified posting here hahah).
CormJack
CC BY-SA 4.0
2023-04-29T11:09:02.133Z
@ Cormjack Just another observation: the geomentric intuition follows from the definition of *definite* integral, as the result is that the area under the graph of a *positive*  function is  the definite integral. But this requires the definition of definite integral and the definition of *what is an area*. Both are not trivial.
BakerStreet
CC BY-SA 4.0
2023-04-29T11:13:40.877Z
@ Cormjack If you are interested and have time, look at these definitions on a book of mathematical analysis, they are not only fundamental, and not trivial, but also fascinating. You may also discover that also infinite parts of the plane have finite area.
BakerStreet
CC BY-SA 4.0
2023-04-29T11:15:42.293Z
If you're talking about the definition of the Riemann integral using our limits of sums when we partition our upper and lower estimates of an area into the sum of smaller and smaller segments. The mechanics of this I believe I understand, and I know this leads to the FTC. But I just don't think I fully appreciated the "fundamentallness" or the subtitles you have mentioned at the time.
CormJack
CC BY-SA 4.0
2023-04-29T11:17:34.063Z
Hmmm I never thought about an infinite plane having a finite area, but looking at convergence the course did illustrate how some integrals with bounds to infinity converge to a finite value (i.e. area) where as some don't. This must be close to what you are referring to!? And it looked at our varying tests of convergence etc. for these "improper" integrals.
CormJack
CC BY-SA 4.0
2023-04-29T11:20:47.037Z
Exactly that, improper integrals.
BakerStreet
CC BY-SA 4.0
2023-04-29T11:21:59.687Z
Amazing, okay I'm on the right track! I'll keep you posted.
CormJack
CC BY-SA 4.0
2023-04-29T11:24:06.007Z
The subtitles are linked to the mathematical and theorical aspect, not to calculations and applications. For instance the existence of the definite integral is linked to the properties and axioms of real numbers, and the definition of area to the notion of convergence, but these are really theorical aspects.
BakerStreet
CC BY-SA 4.0
2023-04-29T11:26:43.980Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/145670/discussion-between-cormjack-and-bakerstreet).
CormJack
<mathematical-economics>
Geometric Intuition for missing constant of integration
CormJack
# Geometric Intuition for missing constant of integration

(Asked by: username_0 on 2023-04-29)

Hi i want to understand the geometric intuition for why there is no constant of integration with definite integrals.

The Algebraic intuition is fine.

$\begin{align} \int\_0^ax^3\,\mathrm{d}x &=\left(\frac14a^4+C\right)- \left(\frac140^4+C\right)=\frac14a^4\end{align}$

**Question**

If our integral is giving us the area under a curve, then surely the original function $f(x)$ has a different area under it's curve than some other related function $f(x) + C$ (Diagram attached at the end). Therefore even when integrating over a range, we would still need to think about an arbitrary constant C.

**Solution:**

I think I need to separate the interpretations of an integral. Undoing differentiation (returning us to our original function). Vs giving us the area under a curve.

1. **Indefinite Integral:** This is getting us back to the function that was differentiated, which will have some arbitrary constant C somewhere. The C does not represent extra potential area, just the possible lost constant when differentiating.
2. **Definite integral**: Is literally asking for the area under that specific function (The integrand) that is being integrated, within that range of integration. Hence no arbitrary constant C is needed?

Although splitting my understanding of an integral into these alternate but related interpretations doesn't feel quite right. Thoughts?

[![enter image description here](https://i.sstatic.net/s3Gia.jpg)](https://i.sstatic.net/s3Gia.jpg)

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: Your observations are very keen, but you are re-discovering the Fundamental Theorem of Integral Calculus: you are in good company, with excellent mathematicians as Newton and Leibnitz. The fundamental Theorem of Integral  Calculus links definite integral with indefinite integral/derivative, and it is really fundamental. It is impossible to answer this question without reporting here some pages of  a mathematical analysis book.  
> **username_2**: This playlist has some videos on integration which I believe can answer your question: https://youtube.com/playlist?list=PLUJGfL_499TI4ihvXuiI8CFUbpp4BawdP  
> **username_1**: @ Cormjack To have an idea, only think  that the *definition* of *definite* integral has nothing to do with the formula you wrote at the beginning of your question, and nothing to do with differntiation. The link with differentiation is given subsequently, precisely with the Fundamental Theorem.  
> **username_0**: Ahh great to see you both as usual, thanks for the comments.  
> **username_1**: @ You are welcome!  
> **username_0**: Hahah yes BakerStreet, I feel like self teaching this stuff, is just permanent rediscovery of things people solved a billion years ago. Your observation is actually very astute, because I remember when I took may advanced calculus class, and I learnt all about integration from first principles and the fundamental theorem of calculus etc. I remember being able to do the mechanics, but not really getting the point of it. I felt at the time it was just telling me things I already new. I.e. undoing differentiation and an area under a curve. But now I think I'm realising some subtleties.  
> **username_0**: Like I didn't really get what was so fundamental about it. I think now if I review that material and the playlist Amit has provided I wil appreciate it more.  
> **username_0**: Yes your last comment BakerStreet is what I was thinking. I will try review this material soon and if I have time post an answer here. (I have another integration question I'm going to post today as well...but it has more economics in it so is a bit more justified posting here hahah).  
> **username_1**: @ Cormjack Just another observation: the geomentric intuition follows from the definition of *definite* integral, as the result is that the area under the graph of a *positive*  function is  the definite integral. But this requires the definition of definite integral and the definition of *what is an area*. Both are not trivial.  
> **username_1**: @ Cormjack If you are interested and have time, look at these definitions on a book of mathematical analysis, they are not only fundamental, and not trivial, but also fascinating. You may also discover that also infinite parts of the plane have finite area.  
> **username_0**: If you're talking about the definition of the Riemann integral using our limits of sums when we partition our upper and lower estimates of an area into the sum of smaller and smaller segments. The mechanics of this I believe I understand, and I know this leads to the FTC. But I just don't think I fully appreciated the "fundamentallness" or the subtitles you have mentioned at the time.  
> **username_0**: Hmmm I never thought about an infinite plane having a finite area, but looking at convergence the course did illustrate how some integrals with bounds to infinity converge to a finite value (i.e. area) where as some don't. This must be close to what you are referring to!? And it looked at our varying tests of convergence etc. for these "improper" integrals.  
> **username_1**: Exactly that, improper integrals.  
> **username_0**: Amazing, okay I'm on the right track! I'll keep you posted.  
> **username_1**: The subtitles are linked to the mathematical and theorical aspect, not to calculations and applications. For instance the existence of the definite integral is linked to the properties and axioms of real numbers, and the definition of area to the notion of convergence, but these are really theorical aspects.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/145670/discussion-between-cormjack-and-bakerstreet).  

---

## 0 answers

