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Suppose f : [a; b… - QuestionCove
OpenStudy (anonymous):

Suppose f : [a; b] to R is Riemann integrable. Then the function F(x) = integral (a to x) f is differentiable on (a; b) and F'(x) = f(x) for all x within (a; b). This is false according to the solution, but I cannot see why, can some1 explain?

6 years ago
OpenStudy (anonymous):

sorry. i would have said yes. must be something subtle

6 years ago
OpenStudy (anonymous):

i guess this is correct only .

6 years ago
OpenStudy (anonymous):

ooh wait

6 years ago
OpenStudy (anonymous):

it is possible for a function to be reimman integrable and still have discontinuities

6 years ago
OpenStudy (anonymous):

in fact there could be just places where the function is not defined at all

6 years ago
OpenStudy (anonymous):

yea this formula u put up is valid for continuos and differentiable f(x)...satellite is rite.

6 years ago
OpenStudy (anonymous):

countably many if i remember correctly. so if f(x) does not even exist at c in the interval (a,b) i guess it wouldn't make sense to say F'(c)=f(c)

6 years ago
OpenStudy (anonymous):

if it is discontinuous at finitely many points than it is still R. integrable

6 years ago
OpenStudy (anonymous):

finitely, and in fact i think it is countable.

6 years ago
OpenStudy (anonymous):

i mean the criteria is "countably many" discontinuities

6 years ago
OpenStudy (anonymous):

thanks, I get it now

6 years ago
OpenStudy (anonymous):

riemann integral is essentially a summation of infinitesimals. so wen there is a discontinuity,u jump over those discontinuities and continue integration over the interval.

6 years ago
OpenStudy (anonymous):

got it! too early to think of such things

6 years ago
OpenStudy (anonymous):

there is another where I have doubts

6 years ago
OpenStudy (anonymous):

f is continuous at \[x _{0} \in(a,b)\] if for every D > 0 there exists E> 0 such that \[\left| x-x _{0} \right| <D\] implies \[\left| f(x)-f(x _{0)} \right|<E\]

6 years ago
OpenStudy (anonymous):

D is delta, E is epsilon And this is also false, but cannot see why

6 years ago
OpenStudy (anonymous):

Oh I found it by googeling, it is swapped, the correct is for every E>0 there exist D>0

6 years ago
OpenStudy (anonymous):

trick stylee

6 years ago
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