Definition of the Definite Integral

  • #1
In the Fundamental Theorem of Calculus, it is stated that

lim max deltax_k -> 0 sigma f(x_k*)*deltax_k = L,

where L is the definite integral of f(x) on [a,b].

How do you apply this limit definition to a normal function to find the definite integral? (Could someone give me math examples of this application?)

Is it like the method used with left and right endpoints in the limit definition where n -> oo for area under a curve? How is it different?

Thanks.
 
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Answers and Replies

  • #2
HallsofIvy
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Soaring Crane said:
In the Fundamental Theorem of Calculus, it is stated that
lim max deltax_k -> 0 sigma f(x_k*)*deltax_k = L,
where L is the definite integral of f(x) on [a,b].
How do you apply this limit definition to a normal function to find the definite integral? (Could someone give me math examples of this application?)
Is it like the method used with left and right endpoints in the limit definition where n -> oo for area under a curve? How is it different?
Thanks.
Quite frankly, I have never seen anything like that in the Fundamental Theorem of Calclus!
There are two parts to the Fundamental Theorem:
1) If [itex]F(x)= \int_a^x f(x) dx[/itex], then F'(x)= f(x).
2) If F'(x)= f(x) then [tex]\int_a^b f(x)dx= F(b)- F(a)[/tex]

Neither of those include the definition of the definite integral which is how you titled this thread. I'm not impressed by the formula you give- it leaves too much unsaid. What exactly do you mean by "deltax_k"? How is x_k* defined?

I THINK what you mean is that you partion the interval from, say, x= a to x= b, into n pieces such that the length of kth "piece" is deltax_k. Choose x_k* to be a value of x in that kth "piece" and sum the products
f(x_k*)deltax_k over k. But for that to be used, you have to give a rule for finding each delta_k not only for each k but for all values of n, as n goes to infinity. I also am not clear what "lim max" means. Do you mean "lim sup"?

The hard part of showing that a function is integrable is showing that none of those choices affect the final result!

An example that is given in just about every calculus book is this:
Let f(x)= 2x+ 3, a= 0, b= 1. For each n, take all deltax_k= 1/n so that each interval has the same length. Of course, the points separating each interval are x_k= k(deltax_k)= k/n. That is, x_0= 0, the left endpoint, x_1= 1/n, x_2= 2/n,..., up to x_n= n/n= 1, the right end point.
For each interval, take x_k* to be the right endpoint of that interval:
x_k= k/n for k= 1 to n. Then f(x_k*)= 2k/n+ 3. The "Riemann sum" for that n is
[tex]\sum_{k=1}^n \left(\frac{2k}{n}+ 3\right)\frac{1}{n}[/tex]
We can separate that into two sums:
[tex]\frac{2}{n^2}\sum_{k=1}^n k+ \frac{3}{n}\sum_{k=1}^n 1[/tex]
It is well known that
[tex]\sum_{k= 1}^n k= \frac{n(n+1)}{2}[/tex]
and, of course,
[tex]\sum_{k=1}^n 1= n[/tex].

That means that, for each n, the sum is
[tex]\frac{n(n+1)}{n^2}+ \frac{3n}{n}[/tex]
The limit of that, as n goes to infinity, is 1+ 3= 4.

Of course, the anti-derivative of 2x+ 3 is x2+ 3x which, evaluated at x=1 and x= 0 gives 4 and 0, which, subracting gives 4 as the value of the integral. That's a lot easier!

By the way, DesCartes, Fermat, and others knew how to find tangent lines (derivatives) and limits of sums (integrals). It was the fact that Newton and Leibniz recognized the identity of "integrals" and "anti-derivatives" that made them the "founders" of calculus.
 
  • #3
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I think s/he means

[tex]\lim_{\max\Delta x_k\rightarrow 0}~sum=L[/tex]

- Kamataat
 
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  • #4
JasonRox
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HallsofIvy said:
Quite frankly, I have never seen anything like that in the Fundamental Theorem of Calclus!
There are two parts to the Fundamental Theorem:
1) If [itex]F(x)= \int_a^x f(x) dx[/itex], then F'(x)= f(x).
2) If F'(x)= f(x) then [tex]\int_a^b f(x)dx= F(b)- F(a)[/tex]

Neither of those include the definition of the definite integral which is how you titled this thread. I'm not impressed by the formula you give- it leaves too much unsaid. What exactly do you mean by "deltax_k"? How is x_k* defined?

I THINK what you mean is that you partion the interval from, say, x= a to x= b, into n pieces such that the length of kth "piece" is deltax_k. Choose x_k* to be a value of x in that kth "piece" and sum the products
f(x_k*)deltax_k over k. But for that to be used, you have to give a rule for finding each delta_k not only for each k but for all values of n, as n goes to infinity. I also am not clear what "lim max" means. Do you mean "lim sup"?

The hard part of showing that a function is integrable is showing that none of those choices affect the final result!

An example that is given in just about every calculus book is this:
Let f(x)= 2x+ 3, a= 0, b= 1. For each n, take all deltax_k= 1/n so that each interval has the same length. Of course, the points separating each interval are x_k= k(deltax_k)= k/n. That is, x_0= 0, the left endpoint, x_1= 1/n, x_2= 2/n,..., up to x_n= n/n= 1, the right end point.
For each interval, take x_k* to be the right endpoint of that interval:
x_k= k/n for k= 1 to n. Then f(x_k*)= 2k/n+ 3. The "Riemann sum" for that n is
[tex]\sum_{k=1}^n \left(\frac{2k}{n}+ 3\right)\frac{1}{n}[/tex]
We can separate that into two sums:
[tex]\frac{2}{n^2}\sum_{k=1}^n k+ \frac{3}{n}\sum_{k=1}^n 1[/tex]
It is well known that
[tex]\sum_{k= 1}^n k= \frac{n(n+1)}{2}[/tex]
and, of course,
[tex]\sum_{k=1}^n 1= n[/tex].

That means that, for each n, the sum is
[tex]\frac{n(n+1)}{n^2}+ \frac{3n}{n}[/tex]
The limit of that, as n goes to infinity, is 1+ 3= 4.

Of course, the anti-derivative of 2x+ 3 is x2+ 3x which, evaluated at x=1 and x= 0 gives 4 and 0, which, subracting gives 4 as the value of the integral. That's a lot easier!

By the way, DesCartes, Fermat, and others knew how to find tangent lines (derivatives) and limits of sums (integrals). It was the fact that Newton and Leibniz recognized the identity of "integrals" and "anti-derivatives" that made them the "founders" of calculus.
Well, we can't say that Fermat didn't recognize this, he might've not considered it important! :eek:
 
  • #5
Hurkyl
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How do you apply this limit definition to a normal function to find the definite integral? (Could someone give me math examples of this application?)
Your calc book ought to have examples, but I haven't done one of these in ages, so I want to give an example!

First off, I would like to emphasize that this is not how one does integrals! It's not necessarily how one should even think about integrals! (But it does have some intuitive appeal in certain circumstances) The point of the limit of Riemann sums definition is that it merely provides some definition from which we can prove things. (And, it is a reasonably constructive definition, which satisfies some philosophical issues)


I want to find the integral [itex]\int_0^1 x^2 \, dx[/itex], if it exists.


So, I write out the limit definition:

[tex]
\lim \sum_{i = 0}^{n-1} (x_i^*)^2 \Delta x_i
[/tex]

(I prefer the convention that [itex]\Delta x_i := x_{i+1} - x_i[/itex])

I first take the subset of partitions that involve equally spaced [itex]x_i[/itex]. So,

[tex]
\lim \sum_{i=0}^{n-1} (x_i^*)^2 \frac{1}{n}
[/tex]

where [itex]i / n \leq x_i^* \leq (i+1) / n[/itex].

I can put bounds on the sum:

[tex]
\frac{(n-1)n(2n-1)}{6n^3} \sum_{i=0}^{n-1} \frac{i^2}{n^3} \leq \sum_{i=0}^{n-1} (x_i^*)^2 \frac{1}{n} \leq \sum_{i=0}^{n-1} \frac{(i+1)^2}{n^3} = \frac{n(n+1)(2n+1)}{6n^3}
[/tex]

and the squeeze theorem shows that when restricting to partitions of equally spaced points, the limit goes to 1/3.


Now, I consider an arbitrary partition. Given any partition into [itex]x_i[/itex]'s, I can subdivide it into something that looks like an equally spaced partition.

In other words, if I have the partition [itex]x_i \, (0 \leq i \leq n)[/itex], I can pick some sufficiently large m and consider the equally spaced partition into equally spaced [itex]y_i \, (0 \leq i \leq m)[/itex].

These two partitions have a greatest common refinement. However, most of the intervals in the greatest common refinement will be exactly the intervals of the [itex]y_i[/itex].


So what I do is that I pick m so that

(1) any Riemann sum over the partition [itex]y_i[/itex] will be within [itex]\epsilon / 2[/itex] of the limit of the equally spaced partitions.

(2) any Riemann sum over the greatest common refinement will be within [itex]\epsilon / 2 [/itex] of a Riemann sum over the [itex]y_i[/itex]'s.


I really don't feel like working out the technical details of that right now. :frown: But once you do, I've proven that the limit exists and is 1/3.
 
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  • #6
Thanks, Kamataat. You wrote the limit more clearly than I did! (I'm sorry that I don't know how to use Latex.) I couldn't find any examples of this limit applied to functions in my textbook, and that is why I am confused. (I do understand the n -> oo limit, however, and its applications to functions.)

Hurkyl, do the partitions have to be equally spaced out using this limit? (Gosh, I feel like I'm asking a question with an obvious answer that I should know! :blushing: )
 
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  • #7
Hurkyl
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Nope. (That's why I talked about the subset of equally spaced partitions, and what to do when the partition is not equally spaced! :smile: )
 
  • #8
Galileo
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By the way, that's exactly what [itex]\lim \limits{\Delta x_k \to 0}[/itex] means.
All the intervals in the partition should go to zero, which happans exactly when the largest interval does.
 

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