Integration-Lower/Upper sum, Partition problem

[SolidSnake]

Homework Statement

Suppose f is increasing and continuous on [a,b]. Show for any partition P,

[tex] \int^b_a f(x)dx - L_f(P) \leq [f(b) - f(a)] \Delta x[/tex]

Homework Equations

Not sure if there are any . but for people unfamiliar with this notation:
[tex] L_f(P) [/tex] = Lower sum for f with the partition P
[tex] U_f(P) [/tex] = Upper sum for f with the partition P

The Attempt at a Solution

If you defined [tex] P = \{x_0, x_1, x_2,...,x_n_-_1, x_n\} [/tex] [tex]a = x_0[/tex] and [tex]b = x_n[/tex]

I also know that
[tex] L_f(P) \leq \int^b_a f(x)dx \leq U_f(P) [/tex]
If i subtract The lower sum from both sides i get,

[tex] 0 \leq \int^b_a f(x)dx - L_f(P) \leq U_f(P) - L_f(P)[/tex]

which gives me the left side of what I'm trying to show. Now the problem arises trying to make the right side look like what it should. When the partitions are of equal size, the lower and upper sum's become a telescoping sum (I know this from a previous question), but in the given case, where the partitions can be of any size i can't find a way to make it look like the right side of the inequality. I've expanded both the lower and upper sums but was unable to cancel anything or find a way to manipulate it to make it look like [tex] [f(b) - f(a)] \Delta x[/tex].

Any help would be great.

 

[mighty2000]

Thank you for your post. I can understand your confusion and I am happy to help you with this problem.

Firstly, we need to recall the definition of a Riemann sum for a function f on the interval [a,b] with a partition P = {x0, x1, x2, ..., xn-1, xn}. The Riemann sum is given by:

R(f,P) = ∑f(xi)(xi+1 - xi)

Now, we know that for any partition P, the lower sum Lf(P) is less than or equal to the integral of f on [a,b] which is also less than or equal to the upper sum Uf(P). Therefore, we can write:

Lf(P) ≤ ∫f(x)dx ≤ Uf(P)

Subtracting Lf(P) from both sides, we get:

0 ≤ ∫f(x)dx - Lf(P) ≤ Uf(P) - Lf(P)

Now, we need to make the right side look like [f(b) - f(a)]Δx. For this, we will use the fact that f is increasing on [a,b]. This means that for any x in [a,b], f(x) ≥ f(a) and f(x) ≤ f(b). Therefore, we can write:

f(a) ≤ f(x) ≤ f(b)

Multiplying both sides by Δx, we get:

f(a)Δx ≤ f(x)Δx ≤ f(b)Δx

Now, we can substitute this in the inequality we obtained above:

0 ≤ ∫f(x)dx - Lf(P) ≤ Uf(P) - Lf(P)
≤ f(b)Δx - f(a)Δx = [f(b) - f(a)]Δx

Hence, we have shown that for any partition P, we have:

0 ≤ ∫f(x)dx - Lf(P) ≤ [f(b) - f(a)]Δx

which is the required result.

I hope this helps you understand the problem better. If you have any further questions, please do not hesitate to ask.