Finding the limit:

@sky#11

this assumption i think can be taken.......

since f(x) is integrable.... so F(x) is differentiable..

this is not always true

Though you guys have solved the problem, let me give the solution that was in my mind. I applied the mean value theorem for the integrals. First of all, transform the first integral by the substitution t=x+h to
\int_{a+h}^{b+h}f(t)\ \mathrm{d}t = \int_{a+h}^{b+h}f(x)\ \mathrm{d}x
owing to the fact that the variable of integration in case of a definite limit is a dummy variable. Now, break it up as
\int_{a+h}^{b+h}f(x)\ \mathrm{d}x = \int_a^{a+h} f(x)\ \mathrm{d}x + \int_{a+h}^b f(x)\ \mathrm{d}x + \int_b^{b+h}f(x)\ \mathrm{d}x -\int_a^{a+h} f(x)\ \mathrm{d}x
where I have added ans subtracted \int_a^{a+h} f(x)\ \mathrm{d}x.
The first two terms on the right combine to give \int_a^b f(x)\ \mathrm{d}x . Hence,
\int_a^b f(x+h)\ \mathrm{d}x -\int_a^b f(x)\ \mathrm{d}x =\int_b^{b+h} f(x)\ \mathrm{d}x - \int_a^{a+h} f(x)\ \mathrm{d}x
Now, the mean value theorem says that \int_a^b f(x)\ \mathrm{d}x is equal to (b-a)f(c) for some c\in[a,b]. Accordingly, we have
\int_b^{b+h} f(x)\ \mathrm{d}x = hf(c)\qquad c\in [b,\,b+h]
and
\int_a^{a+h} f(x)\ \mathrm{d}x = hf(d)\qquad d\in [a,\,a+h]
Hence, we have
\lim_{h\to 0} \int_a^b(f(x+h)-f(x))\ \mathrm{d}x = \lim_{h\to 0} (f(c)-f(d))
But as h→0, c→b and d→a. Hence, we get
\lim_{h\to 0} \int_a^b(f(x+h)-f(x))\ \mathrm{d}x = f(b)-f(a)

\lim_{h\rightarrow0}(1/h)\int_{a}^{b}{f(x+h)-f(x)}=\lim_{h\rightarrow0}1/h*h[{f(a+h)+f(a+2h)+f(a+3h)+...................f(b)+f(b+h)} - {f(a)+f(a+h)+f(a+2h)+f(a+3h)...................f(b)}]
h is cancelled, and all other terms in f(x+h)are subtracted by terms in f(x)
=\lim_{h\rightarrow0} f(b+h)-f(a)

That finally gives f(b)-f(a)
since f(x)is continuous on R

let us not venture far on subhash's soln alone wothout getting a conformation from Kaymant sir ...

bcoz if it turns out to be wrong we are back to square 1

personally i feel any continuous function can represent a derivative (of its antiderivative :D )

well did u all see carefuly??

i told if f(x) after integrating becomes F(x), then obviously F(x) can be differentiated to become f(x).

provided f(x) is integrable..

#18

philip said
mind your terminolgy sky..... that guy has a thread on non integrable funcs to his name

wat do u mean philip[7][7]

buut y shud we consider non-integrable functions ... wen it is given dat it is integrated beween certain limits ?

not everthing is differentiable .... i am 'not everything' :P ~~ even though there is a :P there :P

@sky [3]

i was deleting it because i was assuming something which i wasnt totally convinced with thats all :)

Is the answer f(b) - f(a) ?

ok reposting :)

let ∫f(x)=F(x) ( differentiable)

question becomes

Lt _h→0{F(b+h)-F(b)/h -F(a+h)-F(a)/h}

giving F'(b)-F'(a)

which is f(b)-f(a)

so you ppl think i was rite?

this assumption i think can be taken.......

since f(x) is integrable.... so F(x) is differentiable..

∫f(x)dx=F(x) can be taken to be diffntiable i think
cuz f(x) is continuous

@subhash post it again wats the problem

can we take this assumption:: ∫f(x)dx=F(x) which i take to be differentiable as well?????????[7][7]

can you guys post your solutions...obviously, if you derivatives your solutions are wrong.