Why does the integral operation give the area under the curve of a function? For example, [math]f(x)=2x[/math] over the domain a to b is equal to [math]F(x)=x^2[/math] evaluated at [math]F(b) − F(a)[/math]. Why does the operation [math]\frac{x^{n+1}}{n+1}[/math] give the area under the curve of a given function? This is not intuitive at all. If I take the sum of all the infinitesimally small rectangles underneath the curve of a function over the domain a to b, that is the same as performing the operation [math]\frac{x^{n+1}}{n+1}[/math] on the function and evaluating the function at a and b and subtracting the former from the latter. Not one of my calculus professors have been able to explain this phenomenon without muh fundamental theorem of calculus but all that does is relate the derivative to the integral by saying the integral is defined as the antiderivative. It doesn't explain why it gives the area under the curve of a given function.
Why does the integral operation give the area under the curve of a function? For example...
Other urls found in this thread:
en.wikipedia.org
notendur.hi.is
lmgtfy.com
en.wikipedia.org
math.berkeley.edu
twitter.com
>all functions are polynomials
>hurr durr I can't into babby's first measure theory
not an answer.
If you can follow the intuition behind the definition of derivatives, you can follow the intuition behind the definition of integrals. Don't focus on what the integral turns out to be for a given function, but how you can arrive at the correct method of integration in the first place.
Try to think what the dx is,
and why
[math]\int (2x + x^2)dx[/math] is correct
but
[math]\int 2x + x^2 \,dx[/math] isn't
No, the intuition for a derivative makes a lot more sense. The derivative is the slope of a secant line as the distance between the two points approaches 0, and we use the difference quotient evaluated at [math]\Delta x=0[/math] or as [math]\Delta x\rightarrow 0[/math] to find this. However, using binomial theorem, we can prove that evaluating the difference quotient in this way will be mean that [math]\frac{d}{dx}x^n=\lim _{\Delta x\to 0}\left(nx^{n-1}-\frac{1}{2}n\left(n-1\right)x^{n-2}\Delta x+...+nx\Delta x^{n-2}+\Delta x^{n-1}\right)[/math]. This is formula, known as the power rule, can be directly obtained from the difference quotient. The same cannot be said for integration.
Taylor series
see pic from some Veeky Forums autist
explain it in simple terms.
protip: you can't.
Do you know what a telescoping series is? I'll give an example, using finite numbers.
[math]f(n)=(n+1)^2-n^2[/math]
We could simplify it, but we won't right now because this form is useful to us. It's the analog of the derivative. So now let's try to evaluate this:
[math]f(2)+f(3)+f(4)+f(5)[/math]
What do you get? Well if you're smart about it, you'll see that all of the terms cancel each other out except the end terms, and that's why we call this telescoping by the way.
[math]-2^2 +(5+1)^2[/math]
So here we've basically just reproduced the fundamental theorem of calculus but in the finite way. Hopefully this makes a bit of intuitive sense since the f(n) function could have represented height at points n along the width, which would give you, roughly, the area.
the area under a specific piece of curve within x and dx can if f(x+dx)-f(x) is approximately 0 be given by f(x)dx.
Hence, to get the area under the curve we take the sum of all these minor contributions:
sum(f(x)*dx) from x=a to x=b.
Letting dx-> 0 gives the exact area and WOOPS, that's the definition of an integral.
This is simple.. Are you in high school or something, faggot? First of all, in order to even make reference to anti-derivatives requires stronger conditions on the function being considered. The idea is an extension of finding the area of non-rectangular shapes. I am sure you gladly accept that the area of a rectangle is base*height, and that the area of a right triangle is base*height/2, or that the area of a circle is pi*r^2. Well.. You can't even conclude that the area of a unit circle is pi without some type of limiting procedure involving approximations by simpler shapes. Your confusion results not from the concept, but from the fact that you are unwilling to accept that it's nothing more than a mathematical coincidence that there is a relationship between the degree of a single-variable monomial and the area under beneath its graph on some bounded region. If you want to cry about it, then use geometry to argue it like the Greeks.
en.wikipedia.org
This proves that integral stands for area under the curve. I recommend you not to be such a brainlet in the future.
Sort of.
A measure kind of gives you the "volume" of a set.
So to explain that definition of the integral...
We have these functions called "Simple Functions." These functions map your domain to a finite set of points.
So the image of each of these simple functions can be sort of viewed a finite horizontal line.
What we have under that sum in the definition is a point in the image of a simple function, multiplied by the "volume" of the preimage of the point (i.e. the "volume" of the portion of the domain which is mapped to the point z)
In the special case where your domain is the real numbers, this is essentially taking the "Area" under the simple function.
Then we want to make a set of these sums. One sum for every simple function that is less than the function f we are trying to integrate. We then take the supremum (smallest possible number that is greater than every number in the set) of this set to get our answer. This corresponds to the area under f.
So you come to Veeky Forums because you have no friends irl?
Nah.. The idea is that it is difficult to compute areas when curves are involved, so we approximate the area by simpler shapes: rectangles, triangles, etc.. The polygonal shapes won't have area exactly equal to the region under the curve, but it will be close modulo some approximating error. If the integral exists, then in all cases this approximating error should tend to zero as you measure using finer and finer shapes. This is why we take limits.. Infinity is a theoretical tool.. In reality, we set an error tolerance and say it's close enough and call it a day. Quit pretending to be good at math.. Taking limits doesn't always give the exact area.
5+7+9+11...
if you let dx approach zero then all you get is a value that infinitesimally small. the integral is not sum(f(x)*dx) from a to b. there is no multiplication involved.
I'm in Calc III and I am gonna finish this semester with a B+ actually. Your post made no attempt to answer my fundamental question which is why the integral operation is the area under the curve.
Stop trying to sound smart. If you can take the limit of a shrinking secant, you can take the limit of a sum of an increasing count of shrinking rectangles. The fact that the upper and lower bounds are variable does not matter. Follow the fucking elementary definition you have been given and pay attention to what you are doing.
>if you let dx approach zero then all you get is a value that infinitesimally small.
Which is why you sum infinitely many of these slices to result in a finite value, which is why we use an integral symbol instead of a sigma.
>there is no multiplication involved.
Yes there is. You can also think of it in the other direction - the infinitesimal dx normalizes the otherwise clearly infinite sum.
Actually, Calc III makes even less sense when it comes to integration. We define the line integral of a vector field as [math]\int \int _Q\left(\vec{F}\left(t\right)\cdot \:d\vec{r}\right)[/math] where Q is the path and [math]\vec{F}\left(t\right)[/math] is a vector field parameterized at t, but in order to evaluate it, you have to take the derivative of a vector-valued function [math]\vec{r}\left(t\right)[/math] and take the dot product of that function and the vector field [math]\vec{F}\left(t\right)[/math]. Only then do you take the integral with respect to the dt parameter. Again, this shit is unintuitive as fuck.
Taking the limit of a shrinking secant is done by evaluating the difference quotient of that function. There is no equivalent to the difference quotient for integration.
[math]\int \underline{ }dx[/math] is just an operation. You don't multiply what you are applying the operation to by dx. That really shows how little you know about calculus desu.
I guess you haven't seen this trick before that's key to understanding telescoping series / fundamental theorem of calculus! Your lucky day! :D
So let's define the difference operator [math]\Delta g(n) = g(n+1)-g(n)[/math] and we see that if we have
[math]g(n)=n^2[/math] then this means [math]\Delta g(n) = f(n)[/math] so we have that this is very much like a derivative.
So now let's move on, and sum over this "derivative".
When I said "be smart about it" earlier this is what I meant. Check this out.
[math]f(2)+f(3)+f(4)+f(5)[/math]
[math][(2+1)^2-2^2]+[(3+1)^2-3^2]+[(4+1)^2-4^2]+[(5+1)^2-5^2][/math]
Notice that I have very strictly just plugged in [math]f(n)=(n+1)^2-n^2[/math]
The key observation here is that I'm incrementing by a single step each time 2, 3, 4, 5 and the function has this built in thing of +1 in it. So they end up cancelling each other out, I'll show you.
[math][(2+1)^2-2^2]+[(3+1)^2-3^2]+[(4+1)^2-4^2]+[(5+1)^2-5^2][/math]
[math]-2^2+[(2+1)^2-3^2]+[(3+1)^2-4^2]+[(4+1)^2-5^2]+(5+1)^2[/math]
Make sure you see where all 8 terms shifted to. It's like a zipper in a way, or like dominos. Each of the combination of brackets equals 0 so we are left with only the end terms:
[math]-2^2+(5+1)^2[/math]
Now haven't you seen that when you also integrate you are only plugging in two end points? This is exactly the same thing.
This sort of thing works entirely in general for pretty much any function,
[math]\sum_{n=a}^{b} \Delta g(n) = g(b+1)-g(a)[/math]
okay but I don't see how this relates to the area under a curve.
"Why" it gives the area is because we define the area under the curve of a non-negative continuous function to be the integral. We do this, because integration is the limit of rectangles of known area. We approximate the area under the curve better and better. It's the area because we want it to be.
Polynomials are dense in the continuous bounded functions.
Hi Calc 2 student! Please stop repeating what your lecturer has told you to help you avoid mistakes. There are many valid abstractions for the same concept. This is one. It makes integrals much more analogous to regular sums.
>It's the area because it converges to the area
>It's the area because we want it to be
Why did you have to ruin a fine response?
Hi idiot! If you had read earlier, I am about to finish Calc III with a B+. Please stop making baseless assumptions to help you avoid mistakes. Not one single person out of the 11 posters has been able to give a thorough answer. I'll rephrase the question in simpler terms. Why does applying the operation [math]\frac{x^{n+1}}{n+1}[/math] give the function to find the area under there curve of [math]x^n{/math] when [math]n\ne -1[/math]?
>Not one single person out of the 11 posters has been able to give a thorough answer.
>we define the area to be this
>here's the operation that determines the area
>no reasoning as to why that particular operation gives the area
>about to finish Calc III with a B+
Then perhaps you should study harder instead of shitposting on Veeky Forums, brainlet.
[math]x^n[/math] when [math]n\ne -1[/math]
Wow, yeah, excuse me for assuming that the area of a rectangle is equal to the product of its two orthogonal sides.
No it doesn't require stronger considerations of the function. Again, we start with an approximation, then, since an infinite area can have any fucking shape we choose, we choose the simpler one.
>muh study harder and you will understand meme
worse than gorillaposting tbhfamalam
still doesn't answer my question. see
int(f(t)*dr)=int(f(t)*dr/dt*dt)
It's just the chain rule.
Go back to the previous post. Each of these f(n) represent heights at a given n, and you're adding them all up... if f(n) was a constant function, then you'd have exactly n*f(n) which is length times width which is area.
You mean that
is OP? Differential equations are going to blow your mind.
Stop rejecting every single explanation at your level given to you and expecting some kind of formal proof without needing any of the higher level math education to understand it.
>I am about to finish Calc III with a B+.
Congratulations on a passing grade, genius. A true rarity.
Yeah I heard of that. Integration is "value accumulation." What I don't understand is why in order to evaluate x^n we have to add 1 to the power of n and divide by n+1 to the the function which allows to find the area for the original function over a given domain.
x^2[math]
>I get what integration means but I'm stuck on a derived and provable result that I can't stare at and reach enlightenment without using the rules of derivatives.
What do you want us to do for you?
>le fundamental theorem of calculus
What am asking, once again, is why the "inverse" power rule gives us the area under the curve. I dunno how to phrase it any differently. I understand the proof for the power rule when it comes to taking the derivative, but not when it comes to integrating.
>implying analysis is just calculus
You are asking for something that can only be shown in terms you do not know. Shut the fuck up and take advanced math, or accept it.
I've given pretty much all the info you need to derive this on your own. It's because the first term of the binomial expansion of x^n is nx^{n-1}.
Sorry but all comes from the fundamental theorem. The proof is not as simple as it is short. You need to read this book, if you study math or physics.
Integration is not defined as an antiderivate. It is defined to be the 'area under a curve' i.e Riemann sum. Fundamental theorem of calculus says you can integrate by finding an antiderivate. You can also integrate by calculating directly the Riemann sum and taking the limit.
The "inverse power rule" gives you the area under the curve because it is the antiderivative. And via the FTC, an integral over an interval of a function is equal to the antiderivative of the function evaluated at the boundary points of said interval.
It is very important to realize that the definition of integration is completely independent of the definition of differentiation. However they are fundamentally related through the FTC.
Okay, so why does the fundamental theorem of calculus say we can integrate by finding the antiderivative? What is the reasoning behind this?
A simple handwaving "proof".
Do you know realize the circular nature of that argument?
>doing this gives you the integral because it is the antiderivative
>the antiderivative is the integral because the FTC says so
Nothing in that proof related the integration to the inverse power rule. I've already said this earlier. Please read the thread.
So basically what I said earlier.
>the inverse power rule is the antiderivative
>the FTC says antiderivative integral
>therefore inverse power rule must give the area under the curve
I don't understand how you could be in Calc 3 and have never seen a proof of the FTC.
Read Rudin :)
The main concept is that F(x_i)-F(x_i-1)+F(x_i-1)-F(x_i-2)+F(x_i-2)-F(x_i-3)...-F(x_i-n)= F(x_i)-F(x_i-n)
You won't get very far if you expect people to spoon feed you. You must work for it to discover the truth on your own. I know how to answer your question, but you've basically just blown me off with a lame attitude. I was showing you the way with finite differences and sums and we could have moved on to deriving [math]nx^{n-1}[/math] simply, but you have put forth nothing. So enjoy spending the next month or probably rest of your life not knowing. lol
Actually this proofs nothing. Only that A'(x)=f(x)
You are confused becouse FTC is so simple. But you should know that it is not an definition and the proof for it is not that simple.
I'm confused at what exactly you want. Are you asking why an integral is equivalent to an antiderivative? Or are you asking why the antiderivative of x^n is (x^(n+1))/(n+1)?
So many people getting trolled hard in this thread.
Actually not. You don't understand why FTC is easy to use but hard to proof.
He's asking for both to be magically explained simultaneously, independent of Calculus and limits entirely.
He is/you are rejecting FTC as even relevant.
Dafuq? It's all about the FTC. Stupid calculus teachers give it without any proof. You need mathematical analysis to proof it.
[eqn] \int_0^b x^n dx = \lim_{N \to \infty} \sum_{i=1}^N \left(\frac{ib}{N} \right)^n \frac{b}{N} [/eqn]
[eqn]= \lim_{N \to \infty} \frac{b^{n+1}}{N^{n+1}} \sum_{i=1}^N i^n[/eqn]
[eqn]= \lim_{N \to \infty} \frac{b^{n+1}}{N^{n+1}} {1 \over n+1} \sum_{k=0}^n (-1)^k{n+1 \choose k} B_k N^{n+1-k} [/eqn]
[eqn]= \frac{b^n}{n+1} [/eqn]
If you know this then it's easy to see that
[eqn] \int_a^b x^n dx = \int_0^b x^n dx - \int_0^a x^n dx = \frac{b^n}{n+1} - \frac{a^n}{n+1} [/eqn]
The other way round, brainlet. It's easeir to prove than to use.
OP here. Nobody has explained yet why the FTC relates the two. I've seen the proof but I still have yet to see why the antiderivative gives the area the curve. As for the user that won't "spoonfeed" me, your logic makes no sense.
Consider a differential operator [math] D:{C^n}\left( \mathbb{R} \right) \to {C^{n - 1}}\left( \mathbb{R} \right) [/math] .
where [math] {C^n}\left( \mathbb{R} \right) [/math] denotes the space of n-times differentiable functions [math] f:\mathbb{R} \to \mathbb{R} [/math].
We want to construct an inverse operator [math] {D^{ - 1}}:{C^{n - 1}}\left( \mathbb{R} \right) \to {C^n}\left( \mathbb{R} \right) [/math] .
Let [math] \[f \in {C^{n-1}}\left( \mathbb{R} \right)\] [/math] be in the image set of the map D.
Then the Fundamental Theorem of calculus tells us [math] \int\limits_a^b {f\left( x \right)\operatorname{dx} } = {D^{ - 1}}\left( f \right)\left( b \right) - {D^{ - 1}}\left( f \right)\left( a \right) [/math].
(proof of the FTC is here: en.wikipedia.org
Where lets say [math] \[\int\limits_a^b {f\left( x \right)\operatorname{dx} } \] [/math] is the riemann integral.
By definition of an inverse, [math] {D^{ - 1}} \circ D = i{\operatorname{d} _{{C^n}\left( \mathbb{R} \right)}} [/math].
So if [math] D\left( f \right)\left( x \right) = n{x^{n - 1}} [/math] then [math] \left( {{D^{ - 1}} \circ D} \right)\left( f \right)\left( x \right) = {x^n} = \frac{{n{x^{\left( {n - 1} \right) + 1}}}}{{\left( {n - 1} \right) + 1}}[/math].
So by the FTC: [math] \int\limits_a^b {{x^m}\operatorname{dx} } = \frac{{{b^{m + 1}}}}{{m + 1}} - \frac{{{a^{m + 1}}}}{{m + 1}}[/math]
>we want to construct an inverse operator
>of an operator which fails to be injective
The integral operation is defined as the area under the curve.
The idea that the integral is the antiderivative evaluated at the endpoints is wrong.
The integral is the antiderivative evaluated at the endpoints only when the antiderivative exists, which is only the case for a VERY limited class of functions.
For more on this, you'll need to look up the definition of the integral and a good proof of the fundamental theorem of calculus.
>all functions are analytic
I forgot how stupid high school math was.
t. someone who has never learned what an integral actually is
OP here. I've looked at several FTC proofs. The first part I understand and is intuitive. The second part is where it gets wishy washy imo. Can someone please clarify this?
>I don't know anything about measure theory
>I don't know anything about the real field of numbers
>I don't know anything about Cauchy sequences
Why do high schoolers try to talk about math again?
This is an important point in your maturation as a mathematician.
We can never prove that a definition matches our intuition.
We have an idea of what the area under a function is. This is our intuition.
We come up with a definition that satisfies our intuition. But insofar as our intuition is not mathematical, we cannot prove that it matches our intuition. Instead, we can check that it satisfies several properties that our intuition requires.
For example, if we define the integral in the usual Riemann way, we find that the integral is linear, the integral from a to b is the same as the integral from a to c plus the integral from c to b, it preserves inequality of functions, it gives the expected area for a rectangle and triangle, etc. None of these facts amount to a proof that the integral is the "area under the function," but it gives us reason to believe that we picked a good definition.
For reading on the Riemann definition of the integral, see the following resource:
notendur.hi.is
pages 129-130.
This method won't work in general.
You can show that this is equivalent to the integral when the integral exists, but the process you outlined may well converge without the integral existing.
It might sound like that means the method you outlined is a better definition because it gives a result even when the integral doesn't. The issue is that it can give a non-unique result. How do we decide which to pick?
If you'd like, I could try to furnish an example.
So that you know what the question you're asking is, that IS what the fundamental theorem of calculus says (for functions that have an antiderivative).
What you should be asking is how the fundamental theorem of calculus is proved.
What's wishy washy? You're literally expecting us to clarify something for you while giving absolutely no context or source to what you're reading. Isn't Veeky Forums 18+?
Just fucking read math.berkeley.edu
Don't get frustrated.
That isn't a circular argument.
The FTC says so is the argument.
What you are wondering now is WHY the FTC says so.
OP. You're really starting to grate people in this thread because you are not trying.
The FTC proves that the antiderivative gives the area under the curve for functions with an antiderivative. The area under the curve is the integral, as defined on pages 129-130 of Rudin's Principles of Mathematical Analysis.
If you want more exposition and more pictures, see chapter 13,14 (I believe) of Spivak's Calculus.
If you want knowledge, you've got to work for it. No one can tell you how you should think about something. Only you can decide what makes intuitive sense and you will only be able to make that decision when you have learned the underlying formalism.
He probably should have done it on the equivalence classes of functions that differ by a constant.
It would have been a bijection then right? (assuming all the differentiability and continuity requirements are met).
>the equivalence classes of functions that differ by a constant.
I didn't bother reading his post because the red errors triggered me but isn't this how every proof of the FTC works? I'm pretty sure it's essential.
> If I take the sum of all the infinitesimally small rectangles underneath the curve of a function over the domain a to b, that is the same as performing the operation x^(n+1)/(n+1) on the function and evaluating the function at a and b and subtracting the former from the latter.
That's the part you don't undertand or what? I don't get your question.
OP. Give me a function such that the derivative of the function is equal to 2x. The ONLY possible function is x^2 + c. What is hard to understand. The power rule for integration is literally just the inverse operations of the power rule for differentiation.
Rudin's proof makes no reference to equivalence classes.
The reason being that the FTC has a more general statement, which rudin does prove.
That is, if F'(x) = g(x) then:
the integral of g from a to b is F(b)-F(a).
None of the posts where I mention this have gotten any attention, so I just want to say it one time clearly so that it's out there for anyone that wants to know.
A lot of people in this thread talk about the integral as being the sum of infinitely many infinitesimal rectangles with height f(x) and width dx.
This gives good intuition for the definition, but it does not always give a unique solution depending on how you choose dx to go to 0.
On the other hand, when the integral exists, this intuitive definition (made only slightly more rigorous) does agree with the value of the integral.
The issue is one of uniqueness. We can't be sure that the sum over all f(x)dx rectangles is unique unless f is already integrable according to some other definition that does guarantee uniqueness.
For example, the function that is 0 on the rationals and 1 on the irrationals does not have a unique sum of this kind. Suppose we wish to find its integral from a to b.
If dx is the distance between consecutive rational numbers, then dx is indeed infinitesimal (it's smaller than any real number you can name). If we start our sum on a rational number, the sum over f(x)dx from a to b is then the sum over 0*dx, which is 0.
Similarly, if dx is the distance between irrational numbers and we start our sum on an irrational number, then the sum over f(x)dx from a to b is the sum over 1*dx, which is 1*(b-a)=b-a.
I hope this is clear. Sorry for the lack of Tex, I'm going to learn how to Tex soon, but my classes have been killing me this quarter.
Oh boy, here comes the faggot with exotic functions.
Did you even read OP's post? He deals only with well behaved functions which of course are Riemann integrable.
I find the Riemann definition of the integral much clearer than the "infinite sum" definition.
The Riemann definition is precise, clear, well motivated, and fully rigorous (Wildberger autists get out). Meanwhile, the "infinite sum" definition complicates proofs, leads to paradoxes, and tacitly invokes measure theory, something high schoolers are usually not ready for.
sum f(x_n)* delta_X_n = (Mean value theorem) sum F(x_n+) - F(x_n-) = F(b) - F(a) (because the middle term is a Telescoping series)
this is of course only giving the idea of the proof. this is actually only true for n -> infinity. Basically the core trick to get from the sum of rectangles times function value is using the mean value theorem. (this assumes that there exists a function F for which F' = f)
>I find the Riemann definition of the integral much clearer than the "infinite sum" definition.
what is the difference?
The difference is that the infinite sum definition is only sure to give a unique answer when the Riemann integral already exists, whereas the Riemann Integral is only ever unique when it exists, by definition.
For the definition, see the first 2 pages of Chapter 6 of Rudin's Principles of Mathematical Analysis.
See theorem 6.7 for a proof that the infinite sum definition agrees with the integral when the integral exists.
>The difference is that the infinite sum definition is only sure to give a unique answer when the Riemann integral already exists
when i studied calculus I what i learned is that the riemann integral is only said to exist if and only if every possible infinite sum converges (then they must converge to the same limit)
so i dont see the difference?!
^^This
Every possible sum converging to the SAME limit might be an equivalent definition.
It would be an interesting exercise to show that it's true if it is. I suspect that it probably is true.
The difference is a matter of formulation. In the Riemann definition, we take infimum over all upper sums (sum from i=1 to n of A_i*(x_i - x_i-1) where A_i>=f(x) for x in [x_i-1, x_i]) and the supremum over all lower sums (defined similarly). If the infimum and supremum are equal, we say that the integral exists and is equal to this shared value. The important point is that each upper and lower sum here is over a finite number of intervals that partition the set [a,b].
That's what I take to be the principle virtue of this definition, that you treat only finite sums (and then take an infimum or supremum), which don't have a lot of the wacky properties that infinite sums can have.
This feels to me like the same idea said more formally using terms that are generally introduced in analysis. The "infinite sum" is convergence of finite sums.
The definitions definitely are distinct.
The idea of every possible infinite sum converging to the same value is fully rigorous and means something different from what I said above.
I just wonder if they are equivalent. My guess is, again, that they are.
If they are, that's pretty cool. I'd still prefer the Riemann definition, but a mathematician should have as many different equivalent definitions in his head at once as possible. That's an important part of the thinking required to formulate concise, meaningful proofs.
its reimann sums. you can find the area under the curve by measuring the area of an arbitrary number of rectangles.
>literally just the inverse operation
Refer to Literally all you have to do is think of "upper" and "lower" integrals summing terms of the form f(x_i)(x_(i+1)-x_i) where f is respectively the max or min over the interval. That these converge to the same values is a necessary and sufficient condition for the existence of the riemann integral
I forget if it was Rudin or Rosenlicht that developed it this way, but it gives a clearer intuitive picture of how abstract functions fail to be riemann integrable
Especially if you're going to have them prove simple continuity is a sufficient condition themselves, since its very easy to intuitively see continuity implying the above
>bump
for stupidity of op
isnt the dx kind of an end cap for the integral? these two are exactly the same imo