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I understand that the integral of any function from and to the same point must equal zero, such; $$\int_a^a f(x) \,dx= 0$$ It makes sense, area from one point to the same point should be zero. But, how is this shown in a mathematical sense, proving it fully, without just saying 'it makes sense'?

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    $\begingroup$ I would say it goes by definition, but it is a matter of taste. $\endgroup$ – mickep May 8 '17 at 6:18
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    $\begingroup$ By definition $$\int_{a} ^{a} f(x) \, dx=0$$ for any function $f$. If $f$ is Riemann integrable on $[a, b] $ then by definition we have $$\int_{b}^{a} f(x) \, dx=-\int_{a} ^{b} f(x) \, dx$$ These definitions are provided so that the following relation holds $$\int_{a} ^{c} f(x) \, dx+\int_{c} ^{b} f(x) \, dx=\int_{a} ^{b} f(x) \, dx$$ holds for any points $a, b, c$ lying in an interval on which $f$ is Riemann integrable irrespective of the linear order of $a, b, c$. $\endgroup$ – Paramanand Singh May 8 '17 at 9:23

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Consider the definition of Riemann Sum on $[a,b]$

$$ \sum_{k=1}^n f(x_j^\ast)(x_j-x_{j-1}) $$

where $a=x_0 \leqslant x_1 \leqslant \dots \leqslant x_{n-1} \leqslant x_n=b$ is a subdivision of $[a,b]$ and $x^\ast_j\in [x_{j-1},x_j]$, $j=1,\dots,n$, are sample points. Then, if $a=b$, we clearly have $a=x_0=x_1=\dots=x_n=b$, so any Riemann Sum is trivially $0$. It converges to $0$, so $\int_a^b f(x)\; dx=0$.

Note: As noted in @Eric Schmidt's answer, this depends a lot on your choice of definition. In the explanation above, I assumed that points in a subdivision could be equal, which is certainly not universal. If you assume that they have to be different (strict inequality $<$ instead of large ones $\leqslant$), then you cannot define $\int_a^a f(x)\; dx$ at all, since the Riemann Sum itself, hence its convergence, would not be defined.

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  • $\begingroup$ Actually do you have any idea why somebody would use the strict inequality? $\endgroup$ – lalala May 9 '17 at 13:35
  • $\begingroup$ @lalala: I looked at Courant, Apostol, and Spivak, and they all use definitions of partition that do not allow for repeated points. $\endgroup$ – Matthew Leingang May 9 '17 at 18:22
  • $\begingroup$ surprising, isnt it. I checked Dieudonne vol 1 and he uses <= (but Riemann sum is only a mere exercise there). $\endgroup$ – lalala May 9 '17 at 18:34
  • $\begingroup$ Baby Rudin also uses the definition with strict inequalities. $\endgroup$ – CJ Dowd May 9 '17 at 22:06
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    $\begingroup$ If one requires all partition points to be different, then a partition of the degenerate interval $[a,a]$ consists of only one point, $x_0 = a$. Then the Riemann sum for that partition is the empty sum $\sum_{k = 1}^0 f(x_k^{\ast})(x_k - x_{k-1})$ (whose value is $0$), it's not undefined. $\endgroup$ – Daniel Fischer Jul 27 '17 at 23:05
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This will depend on the exact definitions used. It is almost certainly the case that one of the following holds:

  1. The statement is true by explicit definition.
  2. The statement follows trivially from the definition. (Technically, option 1 is an instance of this.)
  3. Integrals where both bounds are the same are not defined.

Consulting texts conveniently at hand, I find the following:

Calculus with Analytic Geometry, 6th ed., by R. E. Larson et al., follows option 1.

Calculus and Analytic Geometry, 9th ed., by G. B. Thomas and R. L. Finney (the cover says Thomas' Calculus Alternate Edition) follows option 1.

Elementary Analysis: The Theory of Calculus by K. A. Ross, appears to follow option 3. (The notation $[a, b]$ is explicitly restricted to the case $a < b$ in this book.)

Principles of Mathematical Analysis, 3rd ed., by W. Rudin, appears to follow option 3 (for the Riemann(-Stieltjes) integral), though if you allow $[a, a]$ as a closed interval you could interpret this as option 2. That the integral is zero holds because, in this case, all the points in a partition must be equal, so the differences between two successive points are zero, so the upper and lower sums over the partition are zero, so the upper and lower integrals are zero, so the integral exists and is zero. But I don't think Rudin means to include this possibility, because Theorem 6.12(c) is written to avoid it.

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It's really very simple. In pure mathematical sence:

Suppose, \begin{align*} \int f(x)\ dx&=g(x)+c\\ \implies\int_a^af(x)\ dx&=g(x)+c\big|_{x=a}^{x=a}\\ \implies\int_a^af(x)\ dx&=[g(a)+c]-[g(a)+c]\\ \implies\int_a^af(x)\ dx&=0\end{align*}

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    $\begingroup$ This only works if $f$ is continuous though. $\endgroup$ – Wojowu May 8 '17 at 10:20
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    $\begingroup$ Appealing to the fundamental theorem of calculus to justify what is essentially a convention is a good argument, but it doesn't prove the relation. $\endgroup$ – Matthew Leingang May 8 '17 at 13:55
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    $\begingroup$ I suspect that the downvotes are from people (like me) who think that using the fundamental theorem of calculus to prove this is conceptually wrong. I think the other answers that focus on the definition of the definite integral are better. (I didn't downvote.) $\endgroup$ – Ethan Bolker May 8 '17 at 16:25
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    $\begingroup$ The Fundamental Theorem of Calculus depends on the notion of Riemann sum, which is what the OP is ultimately asking about. You've got the cart before the horse. -1. $\endgroup$ – Alfred Yerger May 8 '17 at 19:45
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    $\begingroup$ @MarkMcClure: I think we agree that the whole answer is bad. I said it was a “good” argument, not because I consider it valid, but because it seems convincing on its face. It's kind of like using an integral to compute the circumference of a circle. It looks satisfying, but it's based on relations which depend on the definition of $\pi$ as the ratio of the circumference of a circle to its diameter. $\endgroup$ – Matthew Leingang May 9 '17 at 18:51
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$$\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx$$ And if you set $b=c$, you get $$\int_a^b f(x) dx = \int_a^b f(x) dx + \int_b^b f(x) dx$$ And therefore $$\int_b^b f(x) dx = 0$$

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  • $\begingroup$ I like this argument, but at least somewhere in this process something must be proven from the Riemann sum definition. Not that there's anything wrong with this. $\endgroup$ – Alfred Yerger May 8 '17 at 19:46
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Paramand Singh's comment comes closest to answering this question so far: The relation is not a statement that can be proved, but a convention that exists in order to fit in with some other true statements. It reminds me a little bit of this endless “Why is $0! = 1$?” question on Quora.

For instance, If $a < c < b$ and $f$ is integrable on $[a,b]$, then $$ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx \tag{1} $$ This can be proven from the Riemann sum definition of the definite integral. The definition $$ \int_b^a f(x)\,dx = - \int_a^b f(x)\,dx \tag{2} $$ is consistent with the definition, since when you try to partition an interval $[a,b]$ in the opposite order, the differences $\Delta x_i = x_{i} - x_{i-1}$ become negative, and the Riemann sums flip signs.

I looked at Spivak, Courant, and Apostol, and all state (1) as a theorem, make a moral justification for (2) as a definition, and finally define $$\int_a^a f(x)\,dx = 0 \tag{3}$$ These two definitions allow the theorem (1) to be extended to all triples $(a,b,c)$, including $c<a$, $c=a$, $c=b$, and $c>b$.

Another important fact about the integral is that it makes integrable functions into continuous functions. If $f$ is integrable on $[a,b]$, and $F(x) = \int_a^c f(t)\,dt$, then $F$ is continuous on $(a,b)$. If we use definitions (2) and (3), we can extend the domain of $F$ to the domain of $f$, including $x=a$, $x<a$, $x=b$, and $x>b$. The definitions ensure that the extended function $F$ remains continuous.

It's been said that the best definitions are the ones which make the theorems easy to state and prove. This is a good example.

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Suppose $\int^a_a{f(x) dx}$ exists (if it doesn't, then this question doesn't make much sense). From integration rules ($\int^b_a{f(x) dx} = -\int^a_b{f(x) dx}$), we know that the following holds:

$$\int^a_a{f(x) dx} = -\int^a_a{f(x) dx}$$

Now, there is exactly one value that is its own additive inverse: $0$. Therefore:

$$\int^a_a{f(x) dx} = -\int^a_a{f(x) dx} = 0$$

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  • $\begingroup$ The only thing I worry about is the conditions we required on our proof of $\int_a^b f dx = -\int_b^a f dx$, such as what conditions we place on $f$ (continuous? Integrable?) and whether we required $b>a$ during our proof $\endgroup$ – Brevan Ellefsen May 8 '17 at 14:53
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    $\begingroup$ @BrevanEllefsen The requirements are that $f$ is continuous and integrable on $[a,b]$ (which are implied by assuming $\int^b_a{f(x) dx}$ exists). Then we substitute $b = a$. The requirement that $b > a$ is implied by the previous requirements (since $[a,b]$ is an empty interval if $b < a$). So, it suffices to assume that $\int^a_a{f(x) dx}$ exists. $\endgroup$ – Mego May 8 '17 at 15:01
  • $\begingroup$ @Mego I would say that $\displaystyle \int^a_a f(x) \ \mathrm dx$ always exists for any function $f:\Bbb R \to \Bbb R$. $\endgroup$ – Kenny Lau May 11 '17 at 11:10
  • $\begingroup$ @KennyLau That requires that $f(a)$ is defined. $\endgroup$ – Mego May 11 '17 at 11:11
  • $\begingroup$ @Mego The defined-ness of $f(a)$ follows from the fact that $f$ is a function from $\Bbb R$ to $\Bbb R$. $\endgroup$ – Kenny Lau May 11 '17 at 11:11
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May be, you could consider $$I=\int_a^{a+\epsilon} f(x) \,dx$$ and let $F(x)$ to be the antiderivative of $f(x)$.

So $$I=\int_a^{a+\epsilon} f(x) \,dx=F({a+\epsilon})-F(a)$$ Now, use Taylor series around $\epsilon=0$ to get $$F({a+\epsilon})=F(a)+\epsilon F'(a)+\frac{1}{2} \epsilon ^2 F''(a)+O\left(\epsilon ^3\right)$$ which makes $$I=\epsilon F'(a)+\frac{1}{2} \epsilon ^2 F''(a)+O\left(\epsilon ^3\right)=\epsilon f(a)+\frac{1}{2} \epsilon ^2 f'(a)+O\left(\epsilon ^3\right)$$ and look at the limit of $I$ when $\epsilon \to 0$.

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    $\begingroup$ Only works if $f$ has an antiderivative. $\endgroup$ – Wojowu May 8 '17 at 10:25
  • $\begingroup$ @Wojowu. I totally agree with you ! My mistake since I did not precise this point. $\endgroup$ – Claude Leibovici May 8 '17 at 10:27
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The Fundamental Theorem of Calculus provides a proof.

Also, you are asking for the area of a line, which is zero.

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  • $\begingroup$ Thank you! I understand this, I just didn't think this was the right way to express it - it's a component of a question I was expecting to be a lot harder. Is it really that simple? $\endgroup$ – Eva May 8 '17 at 6:19
  • $\begingroup$ you might be confusing with en.wikipedia.org/wiki/Residue_theorem $\endgroup$ – JMP May 8 '17 at 6:22
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    $\begingroup$ I don't see anywhere in OP's question that says the function is continuous, so this answer actually only covers a special case of the question. $\endgroup$ – Prince M May 8 '17 at 6:23
  • $\begingroup$ on $[a,a]$ continuity isn't relevant $\endgroup$ – JMP May 8 '17 at 16:11
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    $\begingroup$ On $[a,a]$ the fundamental theorem isn't relevant either. What is an antiderivative of a function defined at one point? $\endgroup$ – Jonas Meyer May 8 '17 at 22:38
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If you use the de facto property that $$\int_a^bf(x)\ dx=-\int_b^af(x)\ dx$$ then it follows that $$\int_a^a f(x)\ dx=-\int_a^af(x)\ dx$$ hence $$\int_a^a f(x)\ dx=0.$$

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I'm no mathematician but if a and b are complex there may be more than one path from a to b and back to a. If the complex plane includes singularities then there will be a residue when the resulting area is integrated. This is known as a contour integral, if I remember rightly, and the value of the residue is fixed and does not depend on the path, only on the number of singularities enclosed.

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    $\begingroup$ That is why the notation $\int_a^b f(z)\,dz$ is not good (and not used) unless your domain is simply connected so that there is no path dependence. The notation $\int_{[a,b]}f(z)\,dz$ can less ambiguously indicate the specific line segment path integral. In that case you would have $\int_{[a,a]}f(z)\,dz=0$. $\endgroup$ – Jonas Meyer May 9 '17 at 17:01

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