hiiiiiii
well KAB has told the rite approach .. lemme complete the question ..
0
n [x]dx=
0
1 (0)dx+
1
2 (1)dx+........+
n-1
[n (n-1)dx
I hope u are clear how this can be broken .. rite ..
Now,
0
n [x]dx=
0
1 (0)dx+
1
2 (1)dx+........+
n-1
[n (n-1)dx
= 0 + 1.1 + 2.1 + 3.1 + .... + (n-1).1
= 1 + 2 + 3 + ..... + (n-1)
= n(n-1)/2
I hope the answer is clear ..
cheers