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

yeah! lets get a conformation from kaymant sir first :)

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 )

Y provided??

it is surely provided wen it says :

"wen it is given dat it is integrated beween certain limits!"

sky its ok leave this u were rite

nothing manipal i didnt remember who had posted that post on non integrable functions.....there was a list....
so i mistook it to be u maybe i am getting sleepy...

come to chatbox for discussing this more if u like

#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]

not sure about the solution,

f(x+h)-f(x) is divisible by x+h-x i.e. h, since f is continous.

so now the given integral can be written as

\lim_{h\rightarrow 0}\frac{1}{h}\int_{a}^{b}{hk}

where k=\frac{f(x+h)-f(x)}{h}

therefore limit becomes k(b-a)

which can be writtten as f(b)-f(a)

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

@sky [3]

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

dont fear errors so much ......... rather the unsolved problem! :P

so you ppl think i was rite?

arey subash y did u delete that ? huh!

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

@eureka i took it to be that way i feel im wrong deleting it :)

getting f(b)-f(a)..........hmmm.......is it rite kaymant sir??.........

even me getting same