Question

how to integrate a greatest integer function

Answer 1

see diagram.

Integration of the greatest integer function can be done with definite integrals with limits.  Let us take lower limit as 0. and the upper limit as v.

x = n + m /10ᵃ,
      m/10ᵃ  is the fraction.  n = whole number,  a > 0,  m (integer) >=0
[x] = integer part of x.
x - [x] = fraction part of x.

f(x) =  [x] = greatest integer such that f(x) <= x.

$I=\int \limits_{x=0}^{v} {[x]} \, dx\\\\=\int \limits_{x=0}^{1} {[x]} \, dx+\int \limits_{x=1}^{2} {[x]} \, dx+\int \limits_{x=2}^{3} {[x]} \, dx+....\ \ \ +\int \limits_{x=[v]}^{v} {[x]} \, dx\\\\=\int \limits_0^1 {0} \, dx+\int \limits_1^2 {1} \, dx+\int \limits_2^3 {2} \, dx+....\ \ \ +\int \limits_{x=[v]}^v {[v]} \, dx\\\\=0+1(2-1)+2(3-2)+3(4-3)+....+([v]-1)*1+[v]*(v-[v])\\\\=0+1+2+3+4+....+([v]-1)+[v]*(v-[v])\\\\=\frac{[v]\ ([v]-1)}{2}+[v] (v-[v])\\\\=[v]\frac{(2v-1-[v])}{2}$

If a limit is negative, then we can use the above method to find the answer.

Answer 2

Step-by-step explanation:

the greatest integer function can be integrated by satisfying the value of the limits in the given greatest integer function

e. g.

integeration of [1/x] with limits 1/2 , 1/3

here the given limits are for x

then for 1/x these limits will be 2 and 3

and the greatest integer function lying between 2 and 3 is 2

thus

we will do the integration of 2dx with limits 1/2 and 1/3