1How could define d([x])!!!!!!
12 Answers
1i found this thing in a book.
∫ab f(x)dg(x) + ∫ab g(x)df(x)= g(b)f(b)-g(a)f(a)
substitute f(x)=(x2+1) & g(x)=[x].
thus, ∫0 3 (x2 +1)d[x] + ∫0 3 2x[x]dx = 30-0
or, ∫0 3 (x2 +1)d[x] = 30- ∫0 3 2x[x]dx
or, ∫0 3 (x2 +1)d[x] = 30-13=17
262@aritra- it seems correct but i am doubtful since if we take the limit to be from 0→3- then answer will vary to a large extent
1@aditya- how will the answer vary? if we use the identity in the second line it is correct.
could someone please help to prove the identity?
262d[f(x)g(x)] = f(x).d{g(x)} + g(x).{f(x)}
then integrate with limits on both sides.
i think the answer will vary due to - g(b)f(b)-g(a)f(a)
putting b=3 , g(b)=3 but putting b=3- , g(b)=2
262finally got through !
I = 0∫1 (x2+1) d[x] = 0∫1-h (x2+1)d[x] + 1-h∫1 (x2+1)d[x]
I1 I2
here h→0+
now for I1, d[x] =0 thus I1=0
for I2 , in the range 1-h to 1 , x2+1 being constant comes outside the integral
thus I2 = (12+1) 1-h∫1 d[x]
now d[x] =1 since value of [x] changes abruptly at x=1
thus I2 = 2 * 1 =2
similarly calculating the required integrals from 1to 2 and 2 to 3 ,
we get I (from 0 to 3) = (12+1) + (22+1) +(32+1) = 17
integral (0,3) (x^2+1)d[x] = 17
[x]=x - {x}
d[x]= dx - d{x}
integral (0,3) (x^2+1)d[x]= integral (0,3) (x^2+1)(dx-d{x})
integral (0,3) (x^2+1)d[x]=integral (0,3) (x^2+1)dx - integral (0,3) (x^2+1)d{x}
17=12 - integral (0,3) (x^2+1)d{x}
integral (0,3) (x^2+1)d{x}=- 5
but since (x^2+1) and {x} is positive how is dis continuous summation -ve???
(Note: Ans is 17 and correct...but y dis anomaly?)
62With due respect to all this discussion but are we not talking about non differentiable functions?
and at integer point will d[x] be 1 but the width will be zero so the integral at those integral points will be 0
at non integer points d[x] will be 0 so the integral is bound to be zero!
I am surprised that most of u have an answer to this question... You guys are thinking in the wrong direction!
The solution that Aritra has given wud have been correct if the function was differentiable...
More over integrals of finite functions is continuous... Here limit at 3 will not exist which shows that there is something fundamentally wrong in the proof!
1yeah you are right. however this was a comprehension given in arihant. and that proof was given.
71http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral