Why is f Integrable on [a, b]?

• Mark53
In summary, the function f is integrable on [a, b] if there exists a regular partition of [a, b] into n subintervals such that the maximum and minimum values of f are within the respective subintervals for each i = 1, 2, . . . , n.
Mark53

Homework Statement

Let f [a, b] → R be a non-decreasing function; that is, f(x1) ≤ f(x2) for any x1, x2 ∈ [a, b] with x1 ≤ x2. So f attains a minimum value of m = f(a) and a maximum value of M = f(b) on [a, b]. Let Pn be a regular partition of [a, b] into n subintervals, each of length ∆x = (b − a)/n, and let mi and Mi be the minimum and maximum values of f on the i-th subinterval respectively for each i = 1, 2, . . . , n.

(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1.
(b) Hence show that U(f,Pn) − L(f,Pn) = (Mn − m1) ∆x.
(c) Express (Mn−m1) ∆x in terms of f, a, b, n and use this to explain why f is integrable on [a, b].

The Attempt at a Solution

[/B]
A) this is because Mi must be less than mi as it is a non-decreasing function that is why Mi = mi+1

b)

=Mi∆x-mi∆x

=(Mi-mi)∆x

is this part correct?

c)

(Mn−m1) ∆x

(f(a)-f(b))((b-a)/n)

how do i show that it is integrable from here?

Mark53 said:

Homework Statement

Let f [a, b] → R be a non-decreasing function; that is, f(x1) ≤ f(x2) for any x1, x2 ∈ [a, b] with x1 ≤ x2. So f attains a minimum value of m = f(a) and a maximum value of M = f(b) on [a, b]. Let Pn be a regular partition of [a, b] into n subintervals, each of length ∆x = (b − a)/n, and let mi and Mi be the minimum and maximum values of f on the i-th subinterval respectively for each i = 1, 2, . . . , n.

(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1.
(b) Hence show that U(f,Pn) − L(f,Pn) = (Mn − m1) ∆x.
(c) Express (Mn−m1) ∆x in terms of f, a, b, n and use this to explain why f is integrable on [a, b].

The Attempt at a Solution

[/B]
A) this is because Mi must be less than mi as it is a non-decreasing function that is why Mi = mi+1
This is not an explanation. It's what we call "hand-waving."
Mark53 said:
b)

=Mi∆x-mi∆x

=(Mi-mi)∆x

is this part correct?
No. How does this show that U(f, Pn) - L(f, Pn) = (Mn - m1)∆x?
Also, you started this off with "=Mi∆x-mi∆x". What is on the left side of this equation? Also, what does Mi∆x-mi∆x represent in relation to the function f and the partition?
Mark53 said:
c)

(Mn−m1) ∆x

(f(a)-f(b))((b-a)/n)
?
Mark53 said:
how do i show that it is integrable from here?
How is the term "integrable" defined?

One more thing. This is obviously a calculus problem. Please don't post such problems in the precalculus section.

Mark44 said:
No. How does this show that U(f, Pn) - L(f, Pn) = (Mn - m1)∆x?
Also, you started this off with "=Mi∆x-mi∆x". What is on the left side of this equation? Also, what does Mi∆x-mi∆x represent in relation to the function f and the partition?
One more thing. This is obviously a calculus problem. Please don't post such problems in the precalculus section.

Starting from LHS

U(f,Pn) − L(f,Pn)

=(f(b))*((b − a)/n)-(f(a))*((b − a)/n)

=Mi∆x-mi∆x

=(Mi-mi)∆x

=RHS

Does this show that they are equal to each other now?

Mark53 said:
Starting from LHS

U(f,Pn) − L(f,Pn)

=(f(b))*((b − a)/n)-(f(a))*((b − a)/n)
You are essentially saying here that U(f, Pn) = f(b)Δx and that L(f, Pn) = f(a)Δx. This isn't true.
How are U(f, Pn) and L(f, Pn) defined? You should be using the definitions of these quantiities.
Mark53 said:
=Mi∆x-mi∆x

=(Mi-mi)∆x
This is not the right-hand side. The expression in parentheses is the max. value in the i-th subinterval minus the min. value in the same subinterval. Notice that in the OP, the expression on the right side is (Mn - m1)∆x, not what you have above.
Mark53 said:
=RHS

Does this show that they are equal to each other now?
No.

Mark44 said:
You are essentially saying here that U(f, Pn) = f(b)Δx and that L(f, Pn) = f(a)Δx. This isn't true.
How are U(f, Pn) and L(f, Pn) defined? You should be using the definitions of these quantiities.
This is not the right-hand side. The expression in parentheses is the max. value in the i-th subinterval minus the min. value in the same subinterval. Notice that in the OP, the expression on the right side is (Mn - m1)∆x, not what you have above.

No.
U(f , P) = M1∆x + M2∆x + ··· + Mn∆x.

L(f , P) = m1∆x + m2∆x + ··· + mn∆x

is this correct?

Mark53 said:
U(f , P) = M1∆x + M2∆x + ··· + Mn∆x.

L(f , P) = m1∆x + m2∆x + ··· + mn∆x

is this correct?
Yes. Each one is a sum of n terms. In summation notation, U(f, Pn) is ##\sum_{i = 1}^n M_i\Delta x##, and similar for the lower sum L(f, Pn).

Mark44 said:
Yes. Each one is a sum of n terms. In summation notation, U(f, Pn) is ##\sum_{i = 1}^n M_i\Delta x##, and similar for the lower sum L(f, Pn).
would that mean if we take the max value of U as i=n and the minimum value of L which is i= 1 it will lead to the following:

=Mn∆x-M1∆x

=(Mn-M1)∆x

=RHS

Mark53 said:
would that mean if we take the max value of U as i=n and the minimum value of L which is i= 1 it will lead to the following:

=Mn∆x-M1∆x
No. For one thing, don't start your work off with "=".
More importantly, you need to use the expressions for U and L that you have in post #5, and what you have above isn't what you're supposed to show.

It might be helpful to draw a picture, including the graph of a nondecreasing function, together with your partition.
Mark53 said:
=(Mn-M1)∆x

=RHS

There is still part a) to do...

Mark44 said:
No. For one thing, don't start your work off with "=".
There is still part a) to do...

Does this have something to do with that the function is non decreasing and that the upper sums are greater than the lower sums but by adding one to the lower sum it would make them equal as they would be corresponding to the same value of the function

thus making the statement true: Mi = mi+1 for each i = 1, 2, . . . , n − 1

Mark53 said:
Does this have something to do with that the function is non decreasing and that the upper sums are greater than the lower sums but by adding one to the lower sum it would make them equal as they would be corresponding to the same value of the function
No.

In fact, the statement in part a, as you wrote it, is incorrect, I believe.
Isn't this what it says in your textbook?
(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1

IOW, the right side isn't mi + 1 -- it's mi + 1, the minimum value in the i + 1st subinterval. You do need to know that the function is nondecreasing, however.
Mark53 said:
thus making the statement true: Mi = mi+1 for each i = 1, 2, . . . , n − 1

Mark44 said:
No.

In fact, the statement in part a, as you wrote it, is incorrect, I believe.
Isn't this what it says in your textbook?
(a) Explain why Mi = mi+1 for each i = 1, 2, . . . , n − 1

IOW, the right side isn't mi + 1 -- it's mi + 1, the minimum value in the i + 1st subinterval. You do need to know that the function is nondecreasing, however.

Sorry didn't know how to write the proper notation in my post you are right it should say mi + 1

I am still confused on how i go about explaining it though

Mark53 said:
Sorry didn't know how to write the proper notation in my post you are right it should say mi + 1

I am still confused on how i go about explaining it though
Draw a picture, or better yet, two pictures. A non-decreasing function is one that is either constant (graph is a horizontal line) or strictly increasing (graph rises throughout the interval). Divide the interval into equal subintervals, and locate the minimum and maximum values in each subinterval.

What is the definition of function max and min values?

The function max value is the highest point on a graph or curve that represents the greatest possible output of a given function. The function min value is the lowest point on a graph or curve that represents the smallest possible output of a given function.

How do you find the max and min values of a function?

To find the max and min values of a function, you can use calculus techniques such as differentiation and finding the critical points. The critical points are where the slope of the function is 0 or undefined, which can help identify the max and min values. You can also use a graphing calculator or manually graph the function to visually determine the max and min values.

What are local max and min values?

Local max and min values are points on a graph that represent the highest or lowest point in a small region or interval. These values are only applicable within that specific interval and may not be the overall max or min value of the entire function.

What is the significance of max and min values in real-world applications?

In real-world applications, max and min values can represent important information such as the maximum or minimum amount of a certain resource, the highest or lowest point of a physical structure, or the peak or valley of a periodic function. These values can also help optimize processes and make predictions in various fields such as economics, engineering, and physics.

Can a function have multiple max or min values?

Yes, a function can have multiple max or min values. These values can occur when the function has multiple peaks or valleys, or when the function is not continuous or differentiable. It is important to specify the interval in which the max or min value occurs to accurately represent the function.

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