All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
Floors and decimals integral II (Posted on 2011-07-21) Difficulty: 3 of 5
Each of m and n is a positive integer with m < n. Evaluate this double definite integral in terms of m and n.

[x+y]*{x+y}*(x+y) dx dy for x = 1 to m and, y = 1 to n

*** [x] denotes the smallest integer ≤ x, and {x} is the decimal part of x, ie {x}=x-[x].

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Solution to a lower dimension. | Comment 1 of 3
Instead of using two variables of integration lets just drop it down to one.  Lets integrate [a]*{a}*a da from 1 to n, with n a positive integer.

Since {a} = a-[a] we can substitute this into the expression to be integrated which simplifies to a[a] - a[a]

Now rather than try to integrate the whole discontinuous expression lets just integrate it from n-1 to n.

[a]*{a}*(a) da =
a[a] + a[a] da from n-1 to n
now on this interval [a] is just a constant (n-1)
(n-1) a - a(n-1) da from n-1 to n =
(n-1) (a/3 - a(n-1)/2| from n-1 to n =
(n-1) (n/3 - (n-1)/3 - n(n-1)/2 + (n-1)(n-1)/2) =
(n-1)(n/2 - 1/6) =
(3n - 4n + 1)/6

Now to get from 1 to n just sum:
Σ (3n - 4n + 1)/6 from 1 to n =
(2n - n - n)/12

The two variable solution sought is quite a bit more complicated since x+y can abruptly exceed the next integer as a less predicable time while integrating.  I don't see how a similar method to this would work.

  Posted by Jer on 2011-07-25 00:11:00
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information