Something Must Be Wrong at this Floor

Calculus Level 5

0 2 ( x 2 + 1 ) d x \large \int _{ 0 }^{ 2 }{ ( { x }^{ 2 }+1 ) \, d\lfloor x \rfloor }

Compute the Riemann-Stieltjes integral above.

Notation : \lfloor \cdot \rfloor denotes the floor function .


The answer is 7.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

From Property

a b f ( x ) d ( g ( x ) ) + a b g ( x ) d ( f ( x ) ) = f ( b ) g ( b ) f ( a ) g ( a ) \int _{ a }^{ b }{ f\left( x \right) d\left( g\left( x \right) \right) } +\int _{ a }^{ b }{ g\left( x \right) d\left( f\left( x \right) \right) } =f\left( b \right) g(b)-f\left( a \right) g\left( a \right)

Now Put

f ( x ) = x 2 + 1 a n d g ( x ) = x a = 0 , b = 2 f\left( x \right) ={ x }^{ 2 }+1\quad and\quad g\left( x \right) =\left\lfloor x \right\rfloor \quad a=0\quad ,\quad b=2

0 2 ( x 2 + 1 ) d ( x ) + 0 2 x d ( x 2 + 1 ) = ( 2 2 + 1 ) 2 ( 0 2 + 1 ) 0 \int _{ 0 }^{ 2 }{ \left( { x }^{ 2 }+1 \right) d\left( \left\lfloor x \right\rfloor \right) } +\int _{ 0 }^{ 2 }{ \left\lfloor x \right\rfloor d\left( { x }^{ 2 }+1 \right) } =\left( { 2 }^{ 2 }+1 \right) \left\lfloor 2 \right\rfloor -\left( { 0 }^{ 2 }+1 \right) \left\lfloor 0 \right\rfloor

0 2 ( x 2 + 1 ) d ( x ) + = 10 0 0 2 x d ( x 2 + 1 ) \int _{ 0 }^{ 2 }{ \left( { x }^{ 2 }+1 \right) d\left( \left\lfloor x \right\rfloor \right) } +=10-0-\int _{ 0 }^{ 2 }{ \left\lfloor x \right\rfloor d\left( { x }^{ 2 }+1 \right) }

= 10 0 2 x d ( x 2 + 1 ) = 10 0 2 2 x x d ( x ) =10-\int _{ 0 }^{ 2 }{ \left\lfloor x \right\rfloor d\left( { x }^{ 2 }+1 \right) } \\ =10-\int _{ 0 }^{ 2 }{ 2x\left\lfloor x \right\rfloor d\left( x \right) }

= 10 2 ( 0 1 0. x d ( x ) + 1 2 x d ( x ) ) = 10 2 ( 1 2 x d ( x ) ) =10-2\left( \int _{ 0 }^{ 1 }{ 0.xd\left( x \right) } +\int _{ 1 }^{ 2 }{ xd\left( x \right) } \right) \\ =10-2\left( \int _{ 1 }^{ 2 }{ xd\left( x \right) } \right)

= 10 2 ( 0 1 0. x d ( x ) + 1 2 x d ( x ) ) = 10 2 ( 1 2 x d ( x ) ) = 10 2 ( 2 2 1 2 2 ) = 10 3 = 7 =10-2\left( \int _{ 0 }^{ 1 }{ 0.xd\left( x \right) } +\int _{ 1 }^{ 2 }{ xd\left( x \right) } \right) \\ =10-2\left( \int _{ 1 }^{ 2 }{ xd\left( x \right) } \right) \\ =10-2\left( \frac { { 2 }^{ 2 }-{ 1 }^{ 2 } }{ 2 } \right) \\ =10-3=7

Moderator note:

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 ;)

Otto Bretscher - 5 years ago

Log in to reply

Agreed. The product rule statement is true for differentiable functions, of which x \lfloor x \rfloor is not.

Calvin Lin Staff - 5 years ago

Log in to reply

Actually, the integration by parts can be generalised further. See here

Log in to reply

@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.

Otto Bretscher - 5 years ago

Log in to reply

@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

Rishabh Deep Singh - 5 years ago

Log in to reply

@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.

Otto Bretscher - 5 years ago

@Rishabh Deep Singh @Rishabh Deep Singh : I wrote a problem "inspired by you", pushing your idea a little further. Enjoy!

Otto Bretscher - 5 years ago

Log in to reply

@Otto Bretscher Really Really Thanks Sir.

Rishabh Deep Singh - 5 years ago

@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.

Calvin Lin Staff - 5 years ago

Can you prove the above property used?

Sanchit Ahuja - 5 years ago

Woah! Now this is some nice question I was looking for. Nice thinking. ( + 10 ) \left(+{10}^{\infty}\right)

Ashish Menon - 5 years ago

Log in to reply

Thanks Bro Tomorrow Is My JEE- ADV Paper And i will Post more questions After That.

Rishabh Deep Singh - 5 years ago

Log in to reply

Oh cool, all the best!

Ashish Menon - 5 years ago

How was your paper?

Ashish Menon - 5 years ago

Log in to reply

@Ashish Menon Good second was some difficult.

Rishabh Deep Singh - 5 years ago

Log in to reply

@Rishabh Deep Singh Hope you are in top rank list :)

Ashish Menon - 5 years ago

Log in to reply

@Ashish Menon Lets Hope.

Rishabh Deep Singh - 5 years ago

We have:

0 2 ( x 2 + 1 ) d x = lim n + k = 1 2 n ( x k 2 + 1 ) ( k n k 1 n ) ; x k ( k 1 n ; k n ) = lim n + ( ( x n 2 + 1 ) + ( x 2 n 2 + 1 ) ) = 2 + 5 = 7 \begin{aligned} \int_{0}^{2} (x^2+1)d\left\lfloor x\right\rfloor&=\lim_{n\to +\infty}\sum_{k=1}^{2n} (x_k^2+1)\left(\left\lfloor \dfrac{k}{n}\right\rfloor-\left\lfloor \dfrac{k-1}{n}\right\rfloor\right)&;x_k\in\left(\dfrac{k-1}{n};\dfrac{k}{n}\right)\\ &=\lim_{n\to +\infty}\left((x_n^2+1)+(x_{2n}^2+1)\right)\\ &=2+5=\boxed{7} \end{aligned}

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.

Otto Bretscher - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...