Rogawski 6.2 #60 (Function that D/N Satisfy MVT for Integrals)

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SUMMARY

The discussion focuses on identifying a discontinuous function that fails to satisfy the Mean Value Theorem (MVT) for Integrals. The proposed function, f(x) = {0 if x ≤ 1, 1 if x > 1} on the interval [0,2], demonstrates this failure, as the average value over the interval is 1/2, which does not equal f(c) for any c in [0,2]. Another example provided is f(x) = x for 0 ≤ x < 1 and f(1) = 1000, which also illustrates a discontinuity affecting the MVT outcome.

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Homework Statement


Give an example of a function (necessarily discontinuous) that does not satisfy the conclusion of MVT for Integrals

Homework Equations



MVT for \int = \frac{1}{b-a}\int ^{b}_{a} f(x) dx

The Attempt at a Solution



So I should need one point of discontinuity on every interval that is a subset of the function's domain over [a,b] such that a<b, right?

So couldn't...

f(x) = {0 if x \leq 1, 1 if x > 1} on the interval [0,2].

We have: \frac{1}{2-0}\int^{2}_{0} f(x) dx = 1/2, as f(x) does not equal 1/2.

Would this work?
 
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Your example looks correct.
 
You could do something as simple as changing the value at an endpoint:

f(x)= x if 0\le x&lt; 1, f(1)= 1000.
 

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