Definite integral of step function

[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 $x<k+1$

 

[DEvens]

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.

 

[Ray Vickson]

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 $\int_a^b f(x) \, dx$.

 

[RUber]

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.

 

[WWGD]

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.

 

[SammyS]

...

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 \ . \$