While we were introduced to integration, we were told about some basic concepts that, as we were told, could not be proved based on our level of sophistication. They are as follows:

  1. $$\int_a^b \! f(x) \, \mathrm{d}x=\phi(b)-\phi(a),$$ where $\phi$ is a primitive of $f$ in $[a,b]$
  2. $$\int_b^a \! f(x) \, \mathrm{d}x=-\int_a^b \! f(x) \, \mathrm{d}x$$

When I learned Riemann Integral, I thought I would be able to prove them, as they didn't seem to be too out-of-the-earth type. So, my question is what is the concept of a primitive according to Riemann Integration? And how can it be used to prove $(1)$? If there isn't any, then where should I look?

Also, how can I even conceptualise $(2)$? I mean, Riemann Integration is defined using Sums. How can I Sum in the opposite way? Does it even matter? And how does it become negative?

I'm just a curious high school student, familiar with the basic concepts of Riemann Integration. So, spare me if my questions are too dumb. And it'd be great if you can suggest some study material where I should look for these type of concepts. Thank you in advance.

  • $\begingroup$ 1) is the fundamental theorem of calculus and 2) is usually taken by definition. For 1), look in any text-book on calculus (or search this site for the mentioned theorem). $\endgroup$ – mickep Nov 30 '15 at 6:26
  • $\begingroup$ @mickep And for $(2)$? Is it just taken by definition? There must be some kind of a proof for it. $\endgroup$ – SinTan1729 Nov 30 '15 at 6:32
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    $\begingroup$ (2) is in fact a consequence of (1), i.e. $\int_b^a f(x) \ dx = \phi(a) - \phi(b)$ $\endgroup$ – Siddharth Joshi Nov 30 '15 at 6:33
  • $\begingroup$ @SiddharthJoshi Yes, but this is much like mechanical. I cannot accept an analytical concept that I cannot even visualise. $\endgroup$ – SinTan1729 Nov 30 '15 at 6:35
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    $\begingroup$ No, it is taken by definition. The motivation is that the "usual" integration rules should work... $\endgroup$ – mickep Nov 30 '15 at 6:45

The first question is the fundamental theorem of calculus: it says that if there is a differentiable function $F$ with $\frac{dF}{dx} = f$, and $f$ is Riemann integrable on $[a,b]$, then $\int_{a}^b f(x) dx = F(b) - F(a)$. This can be proved from the definitions relatively easily, but not obviously. It will be covered in any book on elementary real analysis, a good book on calculus, or Wikipedia (NB: When I was a curious high school student, Wikipedia was a godsend for explaining things that didn't make sense or weren't explained in calc class!)

The second statement should be considered a definition. For $a < b$, we define the integral $\int_{a}^{b} f(x) dx$ to be the integral of $f$ with respect to the interval $[a,b]$ (note that the definition of the integral only really depends on the interval!). Then, for any $a,b,c$ with $a < b < c$, we have $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$ (which is easy to see from the definitions). Now, we want to assign meaning to the symbol $\int_b^a f(x) dx$ with $a \leq b$. We can do this in a unique way such that the addition formula above holds for any $a,b,c$. First, plugging in $b = c$, the formula reads $\int_a^b f(x) dx + \int_b^b f(x) dx = \int_a^b f(x) dx$, so $\int_b^b f(x) dx = 0$. Then, if $a < b$, we can plug in $c = a$ to get $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx = 0$, so $\int_b^a f(x) dx = - \int_a^b f(x) dx$.

  • $\begingroup$ Thank you. This makes some sense. $\endgroup$ – SinTan1729 Nov 30 '15 at 6:57
  • $\begingroup$ @Dorebell: in your first sentence, the FToC requires $f $ continuous; Riemann integrable is not enough. $\endgroup$ – Martin Argerami Nov 30 '15 at 9:35
  • $\begingroup$ ah my bad! I thought that differentiability of $F$ was sufficient... $\endgroup$ – Dorebell Nov 30 '15 at 9:58
  • $\begingroup$ @Dorebell: actually, you were right, my bad. I only read "$f $ integrable" and I missed the part "$F $ differentiable". Sorry! $\endgroup$ – Martin Argerami Nov 30 '15 at 10:17
  • $\begingroup$ Riemann integrability of $F' (=f)$ is enough . $\endgroup$ – Tony Piccolo Nov 30 '15 at 12:15

You cannot associate any physical interpretation to $\int_b^a f(x) \ dx$

You're right, the sum would be the same irrespective of whether you sum it forward or backward. But $\int_b^a f(x) dx$ doesn't connote taking the sum in the opposite manner. If it were then $\int_b^a f(x) dx$ would have been equal to $\int_a^bf(a+b-x)dx$ which doesn't happen.

However, this is a useful concept since it allows us to write $\int_a^bf(x) dx = \int_a^cf(x) dx + \int_c^bf(x) dx$ even when $a < b < c$

  • $\begingroup$ Hmm. So, we can say that this has been 'tailored' such that we get desired results. $\endgroup$ – SinTan1729 Nov 30 '15 at 6:53
  • $\begingroup$ yes this is only useful in calculations, nothing else. $\endgroup$ – Siddharth Joshi Nov 30 '15 at 6:54
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    $\begingroup$ I don't agree. In fact, integration is integration of differential form on orientated manifold (here 1-dimentional, so line segment). If you change orientation, sign changes, because we want integration on closed path a → b → a to be zero. Why? it is not obvious on line segments, but it makes much more sense in higher dimentions, and there is very good physical interpretation. For example en.wikipedia.org/wiki/Divergence_theorem $\endgroup$ – user2622016 Nov 30 '15 at 12:48
  • $\begingroup$ Orientation is something different from the selection of initial and final points. For example, if you're talking about line integrals, the line integral is irrespective of the initial and final points of a curve along that path. However, only if you give the same curve with the same parameterization the opposite orientation, the line integral changes sign. More precisely, orientation is for a curve and not for a path. You can associate meaning to line integrals of vector fields although. $\endgroup$ – Siddharth Joshi Nov 30 '15 at 14:30
  • $\begingroup$ Although you can associate notion of orientation of path, or surface to line integrals or surface integrals of vector fields respectively. You cannot do so for scalar fields. $\endgroup$ – Siddharth Joshi Nov 30 '15 at 14:43

$\int_a^b \! f(x) \, \mathrm{d}x=g(b)-g(a)$ is a result obtained directly from the fundamental theorem of calculus.

Say there's a function $f(x)$

Now the area under the curve from $0$ up til $x$ can be found using Reimann sums. Let A(x) = $\int_0^x f(t) dt$ where A(x) is the area function that gives you the area under the curve of $f(x)$ from $0$ up to an $x$ value

It can be easily shown that $A '(x) = f(x)$ (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning)

Which means, $A(x)$ is an antiderivative of $f(x)$ Using this result, we can compute areas under the curve of a function easily. Depending on your choice of antiderivative, you can calculate the area from any point $a$ up to $x$ under the curve. For e.g if you want to find the area from $0$ up to $x$, you will choose an antiderivative g(x) that satisfies the relation g(0) = 0.

Now say you want to find the area from $a$ up to $b$. To make things easier for us, we can choose any antiderivative g(x) we like, evaluate the value of $g(b)$ ie. find the area under the curve up till $b$ and from it subtract $g(a)$ ie the area under the curve up to $a$ which will leave us with the area from $a$ to $b$

The proof for the above can be found here -https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part

  • $\begingroup$ Any input on how I can improve this answer will be appreciated. $\endgroup$ – User2956 Nov 30 '15 at 8:43
  • $\begingroup$ Instead of your approach, which is rather a justification than a proof, the proof for the second part on the same Wikipedia page is useful, as suggested by @Dorebell. $\endgroup$ – SinTan1729 Nov 30 '15 at 9:36
  • $\begingroup$ What I was trying to accomplish with this answer was to provide intuition rather than prove it. Nevertheless, I'll include the proof for the second part too. $\endgroup$ – User2956 Nov 30 '15 at 10:18
  • $\begingroup$ I think we should rather treat this as 'tailored to suit our needs' just as @SiddharthJoshi suggested. $\endgroup$ – SinTan1729 Nov 30 '15 at 10:23

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