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# Integration of greatest integer function

The integration of the greatest integer function, often denoted as [x], where x is a real number, can be a bit tricky because the function is not continuous.
The greatest integer function returns the largest integer less than or equal to x. For example,
[1.2] = 1
[-1.1] = -2
[4.5]=4
[.5]=0
[-.2]=-1
[2]=2

## Integration of greatest integer function

To integrate the greatest integer function from a to b, where a and b are real numbers and a < b, you need to consider the discontinuities that occur at integer values. The integral can be expressed as a sum of integrals over intervals where the function is constant.
Lets see with an example
Example 1
Evaluate
$\int_{1}^{5} [ x ] \, dx$

• Here both the limits are integers
• In this case, the integral can be broken down into the sum of integrals over the intervals [1, 2), [2, 3), [3, 4), and [4, 5). On each of these intervals, the greatest integer function is constant. So, the integral becomes:

$\int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{5} 4 \, dx$

Calculating each integral separately:

$= (2 – 1) \cdot 1 + (3 – 2) \cdot 2 + (4 – 3) \cdot 3 + (5 – 4) \cdot 4$
$= 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4$
$= 1 + 2 + 3 + 4 = 10$

Example 2
Evaluate

$\int_{2.5}^{6.5} [x ] \, dx$

• Here both the limits are non-integers
• This integral can be broken down into the sum of integrals over the intervals [2.5, 3), [3, 4), [4, 5), [5, 6), and [6, 6.5). On each of these intervals, the greatest integer function takes on a constant value. So, the integral becomes:

$\int_{2.5}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{5} 4 \, dx + \int_{5}^{6} 5 \, dx + \int_{6}^{6.5} 6 \, dx$

Calculating each integral separately:

$= (3 – 2.5) \cdot 2 + (4 – 3) \cdot 3 + (5 – 4) \cdot 4 + (6 – 5) \cdot 5 + (6.5 – 6) \cdot 6$
$= 0.5 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 + 1 \cdot 5 + 0.5 \cdot 6$
$= 1 + 3 + 4 + 5 + 3 = 16$

General Formula

Now we have seen some examples, lets see if we can determine some formula based on it

There are four case here
(I)
Let’s say a and b are not integers

If ( n ) is an integer such that ( a < n < b ), the function [x] is constant and equal to ( n – 1 ) for ( x ) in the interval ( [n-1, n) ). Therefore, the integral over this interval is simply the length of the interval times the value of the function.

The integral from a to b of the greatest integer function [x] is:

$\int_{a}^{b} [x] \, dx = \int_{a}^{[a+1]} [x] \, dx + \sum_{n = [a+1]}^{[b] – 1} \int_{n}^{n+1} [x] \, dx + \int_{[b]}^{b} [x] \, dx$

In each interval [n, n+1), the value of [x] is just n, so the integral becomes:

$= [a] \cdot ([a+1] – a) + \sum_{n = [a+1]}^{[b] – 1} n + [b] \cdot (b – [b])$

This formula breaks the integral into parts where the function is constant, and then sums those parts up.

(II) if a and b are integers, the corresponding term simplifies since both ( [a] = a ) and ( [b] = b ).
$\int_{a}^{b} [x] \, dx = \sum_{n = [a+1]}^{[b] – 1} n +a= \sum_{n = [a]}^{[b] – 1} n$

(III)
if a is integer and b is not, the corresponding term simplifies since ( [a] = a ) .

$\int_{a}^{b} [x] \, dx = \sum_{n = [a]}^{[b] – 1} n + [b] \cdot (b – [b])$

(III)
if a is not and b is a inetgers, the corresponding term simplifies since ( [b] = b ) .

$\int_{a}^{b} [x] \, dx = \sum_{n = [a+1]}^{[b] – 1} n + [a] \cdot ([a+1] – a)$

## Solved Examples

Question 1
Find

$\int_{1}^{2} [3x] \, dx$

Solution

let 3x=t
3 dx=dt
dx= dt/3
Therefore

$\int_{1}^{2} [3x] \, dx=\frac {1}{3} \int_{3}^{6} [t] \, dt$
$= \frac {1}{3} [ \int_{3}^{4} 3 \, dt + \int_{4}^{5} 4 \, dt + \int_{5}^{6} 5 \, dt] =4$

Question 2
Find

$\int_{0}^{2\pi} [sin x] \, dx$

Solution

$\int_{0}^{2\pi} [sin x] \, dx = \int_{0}^{\pi} 0 \, dx + \int_{\pi}^{2\pi} -1 \, dx= -\pi$

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