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Let f : Q P be integrable over the n-rectangle Q, and suppose f (c) for all x e Q_ Show that Jo$ (Be careliil: it is possible for m,(f) for subrectangle R € Q...

Question

Let f : Q P be integrable over the n-rectangle Q, and suppose f (c) for all x e Q_ Show that Jo$ (Be careliil: it is possible for m,(f) for subrectangle R € Q evem when /

Let f : Q P be integrable over the n-rectangle Q, and suppose f (c) for all x e Q_ Show that Jo$ (Be careliil: it is possible for m,(f) for subrectangle R € Q evem when /



Answers

Suppose that $f^{\prime}$ is integrable and $\left|f^{\prime}(x)\right| \leq M$ for all $x$. Prove that $|f(x)| \leq|f(a)|+M|x-a|$ for every $a$

Right for this problem we are told that it can be shown with a certain amount of work that if fx is integral on the interval from A to B then so is the absolute value of ffx then asked? Is the converse true? Well, what we can actually do is think of something a little bit like the directly function from the previous problem Where we say OK, let f of x equal one. If x is a rational number And -1. If X is an irrational number, that should have been a queue for rational, this will be an eye the double stroke for irrational. Now, what that means? Based on a similar logic to the previous problem that if we partition or if we evaluate only rationals, we'll get some positive value for the integral. The integral from A to B of fx the x Will be greater than zero. And if we evaluate mhm If we evaluate only on irrational then we'll have that the integral from A to B F x dx will be something less than zero but or the Riemann sum, I should say. So that should be some from I equals one to N of f of x N. A similar deal here Equals one up to end one up to end of fx N D R. Thanks and delta X N. And there should be a delta X in here as well. But if we consider summing over the absolute value, then no matter what we do, we'll get something greater than zero. So that shows the converse is not true. Mhm


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