Proving the Inequality of Upper Darboux Sums for Riemann Integrals

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In summary, the conversation discusses the proof that if a function f is bounded on the interval [a,b] and P and Q are two partitions of this interval, where P is a refinement of Q, then the upper sum of f over Q is less than or equal to the upper sum of f over P. The argument presented is valid for the case where Q refines P by a single additional point, but it can be extended to the general case by considering a sequence of partitions that refine each other by adding one point at a time.
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sandra1
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Homework Statement


Suppose that function f: [a,b] --> R is bounded, and P and Q be 2 partition of [a,b]. Prove that if P is in Q then U(Q,f)<= U(P,f)

Homework Equations


The Attempt at a Solution


P is in Q so suppose there's a c that is in Q but not in P such that c is in between x_i-1 and x_i
M_k = sup{f(x)| x in [x_k-1,x_k]

let's denote:
M_i = sup{f(x)| x in [x_i-1,x_i]
r_1 = sup{f(x)| x in [x_i-1, c]
r_2 = sup{f(x)| x in [c ,x_i]

therefore M_i = max{r_1,r_2}

by definition U(P,f) = [k=1]\Sigma[/n] M_k*(x_k - x_k-1)
= [k=1]\Sigma[/i-1]M_k*(x_k - x_k-1) + M_i(x_i -x_i-1) + [k=i+1]\Sigma[/n]M_k*(x_k - x_k-1)
>= [k=1]\Sigma[/i-1]M_k*(x_k - x_k-1) + r_1(c - x_i-1) + r_2(x_i - c) [k=i+1]\Sigma[/n]M_k*(x_k - x_k-1)
= U(Q,f)
as desired.

Do you see anything wrong with this proof? Thanks very much
 
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  • #2
Your calculation works for the case where [tex]Q[/tex] is a refinement of [tex]P[/tex] by one single additional point [tex]c[/tex]. In general [tex]Q[/tex] refines [tex]P[/tex] by adding some finite number [tex]m[/tex] of points, so you need to observe that the argument can be repeated, say by letting [tex]P \subset P_1 \subset \cdots \subset P_{m-1} \subset Q[/tex] be a sequence of partitions each of which refines the previous by adding a single point.
 
  • #3
ystael said:
Your calculation works for the case where [tex]Q[/tex] is a refinement of [tex]P[/tex] by one single additional point [tex]c[/tex]. In general [tex]Q[/tex] refines [tex]P[/tex] by adding some finite number [tex]m[/tex] of points, so you need to observe that the argument can be repeated, say by letting [tex]P \subset P_1 \subset \cdots \subset P_{m-1} \subset Q[/tex] be a sequence of partitions each of which refines the previous by adding a single point.

yes you're right. i totally forgot about the general case with Q refines P by any finite number of points more than 1. Thanks very much for your response.
 

What is an Upper Darboux Sum?

An Upper Darboux Sum is a type of Riemann Sum used in calculus to approximate the area under a curve. It is found by dividing the area under the curve into smaller rectangles and taking the sum of the areas of the rectangles whose upper edges touch the curve.

What is a Lower Darboux Sum?

A Lower Darboux Sum is similar to an Upper Darboux Sum, but instead of using the upper edges of the rectangles, it uses the lower edges. This means that the rectangles may not fully cover the area under the curve, leading to an underestimate of the total area.

How is an Upper Darboux Sum calculated?

To calculate an Upper Darboux Sum, you first need to divide the interval of the function into smaller subintervals. Then, find the maximum value of the function on each subinterval and multiply it by the width of the subinterval. Finally, take the sum of all these values to get the Upper Darboux Sum.

What is the difference between an Upper and Lower Darboux Sum?

The main difference between an Upper and Lower Darboux Sum is the choice of which edge of the rectangle to use for approximating the area under the curve. The Upper Darboux Sum uses the upper edge, while the Lower Darboux Sum uses the lower edge. This results in different values and can lead to an overestimate or underestimate of the total area.

Why are Darboux Sums important?

Darboux Sums are important in calculus because they allow us to approximate the area under a curve, which is essential in many real-world applications. They also help us to better understand the behavior of functions and their derivatives.

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