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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