# Definite integral of step function

• Byeonggon Lee
In summary: This limit is zero because ##x=k## when it is approached from the left and from the right, and it is also zero when ##x\in[k-1,k]##.
Byeonggon Lee
I need to prove whether this expression is true or false:

##
\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}
##

I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.

In this case, function ##[x]## should be continuous in interval ##[k-1,k]##
but actually [x] is not continuous in ##[k,k-1]##
##[x]=k-1## when ##k-1\leq x<k##, ##[x]=k## when ##x=k##

So I thought that this expression is false,
but in my book's answer, it is true:

"""
Because##[x]=k-1## when ##k-1\leq x<k##, ##[x]=k## when ##x=k##
##
\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx
##
##
=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+...+\int_{n-1}^{n}[x]dx
##
##
=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+...+\int_{n-1}^{n}(n-1)dx
##
##
=0+\bigg[x\bigg]_{1}^{2}+\bigg[2x\bigg]_{2}^{3}+...+\bigg[(n-1)x\bigg]_{n-1}^{n}
##
##
=0+1+2+...+(n-1)
##
##
=\frac{n(n-1)}{2}
##
"""

The book also says that I need to think interval as ##k\leq x<k+1##
and I don't understand how is it possible to omit equal sign at ##x<k+1##

An integral is an area. The troublesome points at the end of the interval mean you are uncertain about the area at a point. What is the difference in the area under a curve if you move one single point up or down by 1?

Another way to approach it is this. Think of the integral as a limit of a sum of areas. It happens that in this case you can very easily construct the limit because the function you are looking at is piece-wise constant.

Byeonggon Lee said:
I need to prove whether this expression is true or false:

##
\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}
##

I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.

The book also says that I need to think interval as ##k\leq x<k+1##
and I don't understand how is it possible to omit equal sign at ##x<k+1##

A definite integral certainly can exist for a discontinuous function. Jump discontinuities in f(x) at an interval's endpoint a or b does not affect the area under the curve y = f(x) from x = a to x = b. In other words,
$$\int_{[a,b]} f(x) \, dx = \int_{(a,b]} f(x) \, dx = \int_{[a,b)} f(x) \, dx = \int_{(a,b)} f(x) \, dx$$
We can denote all four of these by the common symbol ##\int_a^b f(x) \, dx##.

You can read a little more on this idea at https://en.wikipedia.org/?title=Riemann_integral.
For example, to your point on continuity:
"A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)."
Set of measure zero means a finite (or countably infinite) set of distinct points. In the case of the step function is discontinuous at the integers which for a maximum n is finite, and without a maximum is countably infinite. Both sets are measure zero, so you are okay to integrate.

Just to nitpick, or to add a bit: while in this context of step functions discontinuities are finite, you may have in other cases uncountably-infinite sets of measure zero, e.g., the Cantor set.

RUber
Byeonggon Lee said:
...

The book also says that I need to think interval as ##k\leq x<k+1##
and I don't understand how is it possible to omit equal sign at ##x<k+1##
I believe your function is also called the floor function, ##\displaystyle \ \lfloor x\rfloor \,,\ ## and the greatest integer function, ##\displaystyle \ [\![ x]\!]\ .\ ##

On the interval ##\displaystyle \ [k-1\,,\ k)\,,\ ## we have ##\displaystyle \ [x]=k-1\ .##

Take the limit: ##\displaystyle \ \lim_{a\to k^-} \int_{k-1}^a [x]\,dx \ . \ ##

## 1. What is a step function?

A step function is a type of mathematical function that changes its value suddenly at certain points, creating a "step-like" shape. It is defined by a series of constant segments, each with a different value.

## 2. What is the definite integral of a step function?

The definite integral of a step function is the area under the graph of the function between two specific points on the x-axis. It can be calculated by finding the sum of the areas of the individual segments that make up the step function.

## 3. How is the definite integral of a step function different from other types of integrals?

The definite integral of a step function is different from other types of integrals because it only considers the area under the graph within a specific interval. Other types of integrals, such as indefinite integrals, consider the entire area under the graph of a function.

## 4. Can the definite integral of a step function be negative?

Yes, the definite integral of a step function can be negative if the function has segments that lie below the x-axis. In this case, the negative portion of the area will be subtracted from the positive portion, resulting in a negative value.

## 5. How is the definite integral of a step function used in real-world applications?

The definite integral of a step function has many real-world applications, such as calculating the total distance traveled by an object with changing velocity, finding the total profit or loss in a business with fluctuating sales, and estimating the total amount of fluid flowing through a pipe with varying flow rates.

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