∫ 0 2 ( x 2 + 1 ) d ⌊ x ⌋
Compute the Riemann-Stieltjes integral above.
Notation : ⌊ ⋅ ⌋ denotes the floor function .
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As mentioned in the discussion, you have to be very careful when applying any theorem. Always remember to check that the conditions actually hold.
What notion of an integral are you using? It's not our trusted Riemann integral ;)
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Agreed. The product rule statement is true for differentiable functions, of which ⌊ x ⌋ is not.
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Actually, the integration by parts can be generalised further. See here
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@A Former Brilliant Member – I don't think you can generalize the product rule (for derivatives) here, but integration by parts holds for Riemann-Stieltjes integrals if either integral exists.
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@Otto Bretscher – Yeah. Sorry, I meant integration by parts.
Sir But i Define The functions in the Interval from 0 to 1 And From 1 to 2. At which It is Differentiable
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@Rishabh Deep Singh – You need to state that you are using the Riemann-Stieltjes integral , not just a Riemann integral, as Comrade @Pi Han Goh says in his report.
@Rishabh Deep Singh – @Rishabh Deep Singh : I wrote a problem "inspired by you", pushing your idea a little further. Enjoy!
@Rishabh Deep Singh – For the Riemann integral, the function has to be differentiable everywhere. It currently isn't differentiable at the point 2, which is why we cannot apply the Riemann integral.
As mentioned, apparently the Riemann-Stieltjes integral can deal with this issue, but I'm not too familiar with that generalization.
Can you prove the above property used?
Woah! Now this is some nice question I was looking for. Nice thinking. ( + 1 0 ∞ )
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Thanks Bro Tomorrow Is My JEE- ADV Paper And i will Post more questions After That.
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Oh cool, all the best!
How was your paper?
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@Ashish Menon – Good second was some difficult.
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@Rishabh Deep Singh – Hope you are in top rank list :)
We have:
∫ 0 2 ( x 2 + 1 ) d ⌊ x ⌋ = n → + ∞ lim k = 1 ∑ 2 n ( x k 2 + 1 ) ( ⌊ n k ⌋ − ⌊ n k − 1 ⌋ ) = n → + ∞ lim ( ( x n 2 + 1 ) + ( x 2 n 2 + 1 ) ) = 2 + 5 = 7 ; x k ∈ ( n k − 1 ; n k )
How do we know that it suffices to consider an even number of subintervals of equal size?
For this problem your approach would give an answer of 3, which would not be correct.
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From Property
∫ a b f ( x ) d ( g ( x ) ) + ∫ a b g ( x ) d ( f ( x ) ) = f ( b ) g ( b ) − f ( a ) g ( a )
Now Put
f ( x ) = x 2 + 1 a n d g ( x ) = ⌊ x ⌋ a = 0 , b = 2
∫ 0 2 ( x 2 + 1 ) d ( ⌊ x ⌋ ) + ∫ 0 2 ⌊ x ⌋ d ( x 2 + 1 ) = ( 2 2 + 1 ) ⌊ 2 ⌋ − ( 0 2 + 1 ) ⌊ 0 ⌋
∫ 0 2 ( x 2 + 1 ) d ( ⌊ x ⌋ ) + = 1 0 − 0 − ∫ 0 2 ⌊ x ⌋ d ( x 2 + 1 )
= 1 0 − ∫ 0 2 ⌊ x ⌋ d ( x 2 + 1 ) = 1 0 − ∫ 0 2 2 x ⌊ x ⌋ d ( x )
= 1 0 − 2 ( ∫ 0 1 0 . x d ( x ) + ∫ 1 2 x d ( x ) ) = 1 0 − 2 ( ∫ 1 2 x d ( x ) )
= 1 0 − 2 ( ∫ 0 1 0 . x d ( x ) + ∫ 1 2 x d ( x ) ) = 1 0 − 2 ( ∫ 1 2 x d ( x ) ) = 1 0 − 2 ( 2 2 2 − 1 2 ) = 1 0 − 3 = 7