For a real number x let [x] denote the largest integer less than or equal to x and x =x-[x] . The smallest possible integer value of n for which ₁^ n [x] x d x exceeds 2013 is

Step-by-step solution

. ₁^ n [r r d x= _ r=1 ^ n-1 _ r ¹⁺¹ r(x-r) d x _ r=1 ^ n-1 r [ x² 2 -r x ]_ r ^ r+1 _ r=1 ^ n-1 r [ (r+1)²-r² 2 -r .1 ] _ r=1 ^ n-1 r [ 1 2 ]= 1 2 n(n-1) 2 n(n-1) 4 2013n(n-1) 4 2013 (n- 1 2 )² 2013 16+1 4 n 32209 2 + 1 2 least n=91