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One, Two, Three, Floor Integral 2 (Posted on 2022-09-20) Difficulty: 3 of 5
Evaluate:
n
∫ ⌊x*⌊x*⌊x⌋⌋⌋ dx
0
where ⌊p⌋ is the floor function. ie. the greatest integer ≤ p.

For each of n=1, 2, 3, 4.

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution computer solution; computer varification Comment 1 of 1
Using

clearvars,clc
prev=0;
for x=1:.0000001:5
  v=floor(x * floor(x * floor(x)));
  if v~=prev 
    fprintf('%7.5f %4d\n',x,v)
  end
  prev=v;
end

to explore the domain (and a little beyond):

  x     f(x) beginning at
         that x value
1.00000    1
2.00000    8
2.25000    9
2.50000   12
2.60000   13
2.80000   14
3.00000   27
3.11111   28
3.22222   29
3.33333   33
3.40000   34
3.50000   35
3.60000   36
3.66667   40
3.72727   41
3.81818   42
3.90909   43
4.00000   64
4.06250   65
4.12500   66
4.18750   67
4.25000   72
4.29412   73
4.35294   74
4.41176   75
4.47059   76
4.50000   81
4.55556   82
4.61111   83
4.66667   84
4.72222   85
4.75000   90
4.78947   91
4.84211   92
4.89474   93
4.94737   94
5.00000  125

The breaks in value of the function itself appear to be at fractions that are multiples of 1/4, 1/10, 1/9 and 1/11 (up to x=4). Therefore doing this in symbolic (exact) values should be done in increments of 1/1980 to hit all these exactly:

tot=sym(0);
prev=sym(0);prevx=sym(0);
for u=1:4*1980
  x=sym(u/1980);
  v=floor(x * floor(x * floor(x)));
  if v~=prev
    incr=(x-prevx)*(prev);
    tot=tot+incr;
%    disp([x v incr tot eval(tot)])
    fprintf('%7s %3s %6s %12s %17.13f\n',x, v, incr, tot, eval(tot))
    prev=v; prevx=x;
  end
end

resulting in the following table:

        new  prev f(x)     accumulated sum
   x    f(x)  * delta x  exact       decimal
      1   1      0            0   0.0000000000000
      2   8      1            1   1.0000000000000
    9/4   9      2            3   3.0000000000000
    5/2  12    9/4         21/4   5.2500000000000
   13/5  13    6/5       129/20   6.4500000000000
   14/5  14   13/5       181/20   9.0500000000000
      3  27   14/5       237/20  11.8500000000000
   28/9  28      3       297/20  14.8500000000000
   29/9  29   28/9     3233/180  17.9611111111111
   10/3  33   29/9      1271/60  21.1833333333333
   17/5  34   11/5      1403/60  23.3833333333333
    7/2  35   17/5      1607/60  26.7833333333333
   18/5  36    7/2      1817/60  30.2833333333333
   11/3  40   12/5      1961/60  32.6833333333333
  41/11  41  80/33    23171/660  35.1075757575758
  42/11  42  41/11    25631/660  38.8348484848485
  43/11  43  42/11    28151/660  42.6530303030303
      4  64  43/11    30731/660  46.5621212121212

The requested integrals are therefore

      n           integral
               exact     decimal
      1           0   0.0000000000000
      2           1   1.0000000000000
      3      237/20  11.8500000000000
      4   30731/660  46.5621212121212

Wolfram Alpha verifies

integral_0^4 floor(x floor(x floor(x))) dx = 30731/660˜46.5621

  Posted by Charlie on 2022-09-20 10:54:05
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