[PDF] Math 410 Section 6.1: Darboux Sums - Lower and Upper Integrals 1





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Math 432 - Real Analysis II

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Math 410 Section 6.1: Darboux Sums - Lower and Upper Integrals 1

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[PDF] Math 432 - Real Analysis II

define the Upper Darboux Sum of f with respect to P to be U(fP) = (a) With the partition P mentioned above and f(x) = x compute U(fP) and L(fP)



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[PDF] Lecture 33 - MATH 409 Spring 2020 [3mm] Advanced Calculus I

Let f : [a b] ? R be a bounded function Definition The upper Darboux sum (or the upper Riemann sum) of the function f over the partition P is the

define the Upper Darboux Sum of f with respect to P to be. U(f,P) = (a) With the partition P mentioned above and f(x) = x, compute U(f,P) and L(f,P).
  • How do you find the upper Darboux sum?

    For the Upper Darboux sum, since every M(f, [t, t ]) = 1, we end up with a sum of the length of the intermediate subintervals and thus U(f,P) = b ? a. L(f) = sup{L(f,P) P is a partition of [a, b]}. We say that f is (Darboux) integrable over [a, b] if L(f) = U(f).
  • What is the difference between Darboux sum and Riemann sum?

    Answers and Replies
    Darboux worked with lower and upper sums, Riemann with a mean value. There is no essential difference, as e.g. to Lebesgue integrals. Riemann integrals and Darboux integrals have different definitions. However they are equivalent.
  • Use the ? ? P condition to prove that a function is Darboux integrable. Compute the Darboux integral by finding a sequence of partitions Pn such that limn?? U(f,Pn) = limn?? L(f,Pn).
Math 410 Section 6.1: Darboux Sums - Lower and Upper Integrals

1.Introduction:The overall goal of the chapter is to establish the two Fundamental Theorems of

Calculus. The first states that antiderivatives can be used to evaluate integrals and the second states that integrals can be used to construct antiderivatives. It may take a minute or two to remember that antiderivatives and integrals are not the same thing at all and it"s the FTOC which connects them.

2.Upper and Lower Darboux Sums

(a)Partitions:Given a closed interval [a,b], a partition of [a,b] is a division into subintervals. More specifically it is a choice ofa=x0< x1< x2< ... < xn=b. We usually write a partition asP={x0,x1,...,xn}. (b)Darboux Sums:Supposef: [a,b]→Ris bounded andPis a partition of [a,b]. Then the

Lower Darboux Sum is:

L(f,P) =n?

i=1m i(xi-xi-1) wheremi= inf{f(x)|x?[xi-1,xi]} and the Upper Darboux Sum is:

U(f,P) =n?

i=1M i(xi-xi-1) whereMi= sup{f(x)|x?[xi-1,xi]} (c)Note:These are remeniscent of the lower and upper sums encountered in most calculus courses but the use of inf and sup makes them more flexible. (d)Example:A good visual one suffices, especially one in which one of the subintervals has a hole in the function so the inf (or sup) exists but the min (or max) does not. Otherwise it"s just like lower and upper sums. (e)Theorem:Supposef: [a,b]→Ris bounded with lower and upper boundsmandM respectively. Then n i=1m i=1M i=1M(xi-xi-1) (f)Definition:IfPis a partition of [a,b] then a refinement ofPis a partition containing at least thex-values inPand maybe more. In other words we just add more cuts. (g)Theorem (The Refinement Theorem):Supposef: [a,b]→Ris bounded,Pis a partition of [a,b] andP?is a refinement ofP. Then

L(f,P?)≥L(f,P)

and Intuition:The intuition here is that as we refine the partition lower sums go up and upper sums go down.

Proof:Omitted.

(h)Theorem:Supposef: [a,b]→Ris bounded andP1andP2are both partitions of [a,b]. Then Proof:LetP?be the partition obtained by using all thex-values in bothP1andP2 combined. Then

3.Upper and Lower Integrals

(a)Upper and Lower Integrals:Supposef: [a,b]→Ris bounded. Then we define the lower integral offon [a,b] as:? b a f= sup(L) = lub(L) whereL={L(f,P)|Pis a partition of [a,b]} and the upper integral offon [a,b] as: ?b af= inf(U) = glb(U) whereU={U(f,P)|Pis a partition of [a,b]} (b)Theorem:Supposef: [a,b]→Ris bounded. Then b a ?b a f Proof:For any partitionPof [a,b] we knowU(f,P) is greater than or equal to every lower sum soU(f,P) is an upper bound forLand since?b a f= lub(L) we must have?b

However since this is true for allPwe know that?b

a fis a lower bound forUand since ?b af= glb(U) we must have?b a ?b a f (c)Note:Calculating lower and upper Darboux integrals using partitions is extremely difficult because it requires and understanding of what the lower and upper Darboux sums would be for every partition in order to make sense of the required supand inf. (d)Example:Definef: [0,2]→Rbyf(x) = 3. Then?2 0 f= 6 and?2

0f= 6.

(e)Example:Definef: [0,1]→Rbyf(x) = 1 ifx?Qandf(x) = 0 otherwise. Then?1 0 f= 0 and ?1

0f= 1.

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