Question

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?

Answer 1

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

Answer 2

i guess this is correct only .

Answer 3

ooh wait

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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)

Answer 8

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

Answer 9

finitely, and in fact i think it is countable.

Answer 10

i mean the criteria is "countably many" discontinuities

Answer 11

thanks, I get it now

Answer 12

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.

Answer 13

got it! too early to think of such things

Answer 14

there is another where I have doubts

Answer 15

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 \[\left| f(x)-f(x _{0)} \right|<E\]

Answer 16

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

Answer 17

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

Answer 18

trick stylee