For f continuous on \mathbb{R}, find the limit:
\lim_{h\to 0}\dfrac{1}{h}\int_a^b \big(f(x+h)-f(x)\big) \mathrm{d}x
Remember, f has been said only to be continuous and not differentiable.
(rajat, dipanjan, metal: wait for at least one day before you post your solution (if you ever feel like posting)
11@sky#11
this assumption i think can be taken.......
since f(x) is integrable.... so F(x) is differentiable..
this is not always true
66Though 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)
66@subash and Kalyan (Posts #15 and #29 respectively), your solutions are acceptable. However, @b555 (post #21), what do you mean when you say that "f(x+h)-f(x) is divisible by h". What do you think about the divisibility of e-(x+h)2-e-x2 by h?
1\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
1let 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 )
21Y provided??
it is surely provided wen it says :
"wen it is given dat it is integrated beween certain limits!"
1well 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..
1nothing 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
11#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]
39not 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)
1buut y shud we consider non-integrable functions ... wen it is given dat it is integrated beween certain limits ?
11not everthing is differentiable .... i am 'not everything' :P ~~ even though there is a :P there :P
11@sky [3]
i was deleting it because i was assuming something which i wasnt totally convinced with thats all :)
11ok 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)
1this assumption i think can be taken.......
since f(x) is integrable.... so F(x) is differentiable..
1∫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
24can we take this assumption:: ∫f(x)dx=F(x) which i take to be differentiable as well?????????[7][7]
66can you guys post your solutions...obviously, if you derivatives your solutions are wrong.
1did anyone of u used f'(x) in ur solution...!!???
if YES, then ur answer is wrong... [3] (it's not clearly stated in d question whether f is differentiable over [a,b]...!!!)
if NO, then post ur solution...!!! [6]
1getting f(b)-f(a)..........hmmm.......is it rite kaymant sir??.........
