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

Home > Just Math > Calculus
Floors and decimals integral (Posted on 2008-07-06) Difficulty: 2 of 5
Let [x] be the smallest integer less than or equal to x, and let {x} be the decimal part of x, ie {x}=x-[x]. For any integer n>1, evaluate the following integral from 1 to n, in terms of n:

[x]*{x}*x dx

No Solution Yet Submitted by Jonathan Lindgren    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1

Assuming that k is an integer, the integrand in the subinterval k to (k+1) is equal to kx(x-k) = k*x^2 k^2*x.

Integrating this in the said subinterval w.r.t  x, we obtain:

k((k+1)^3 k^3)/3 k^2*((k+1)^2 k^2)/2
= k^2/2 + k/3 (upon simplification)
= f(k) (say)

Then, the required definite integral will be given by:

f(1) + f(2) + + f(n-1)
= (1/2)*Sum (k=1 to n-1) (k^2) + (1/3)*Sum (k=1 to n-1) (k)
= (n(n-1)/12)*(2n+1)
=  n(n-1)(2n+1)/12

                             Q    E    D

Edited on July 14, 2008, 2:32 pm
  Posted by K Sengupta on 2008-07-06 11:58:43

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 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information