Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2011-04-11 17:15:59

GOKILL
Member
Registered: 2010-03-19
Posts: 26

still riemann integral

#2
Given

and

Let

Prove that

i think when the function f is monotonic increasing,that can be done, but otherwise?


I am the greatest magician this century!!!

Offline

#2 2014-05-23 21:43:58

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,436
Website

Re: still riemann integral

3 years late, but might be useful to people looking up the same thing.

First, note that your second partition is a refinement of the first with one extra point -- namely, y. I prove the general case for any two partitions P and P* where P* is a refinement of P, so that your result follows as an immediate corollary.

Let P be a partition of [a,b], with U(f,P) and L(f,P) being the upper and lower Darboux sums of f with respect to the partition P. Write the original partition P as

and the partition P* as

where x* is the extra point. Then:

But note that

and

substituting, we see that the RHS of the inequality is greater than or equal to 0, thus

.

The inequality

is done similarly. Thus, by repeated application of this proof, this holds true for any refinement of P. Using this result, let

be a common refinement of P and Q. Then:

and

but, obviously

hence

QED.

Last edited by zetafunc (2014-05-25 19:12:41)

Offline

Board footer

Powered by FluxBB