# Integral of a constant function with/without endpoint

Flumpster

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## The Attempt at a Solution

I know that given $f(x)=c$, the integral from $a≤x≤b$ is $c(b-a)$ (at least, I hope I know that! :D).
Is the integral the same value if you don't include an endpoint? That is, if you were evaluating $f(x)$ from $a≤x<b$?

Intuitively I think it both integrals should be the same value, for 2 reasons:

1) The two integrals are the same except one of them doesn't include the ordinate of b (does not include $f(b)$ in the set of function values). This can be represented as a vertical line which would have no area.

2) I'm even less sure about this bit, but surely you could use a limit to show that evaluating the integral from $a≤x<b$ - "from a to as close to be as you can get" - is the same as evaluating it up to and including b?

Like I said, I don't really know, and these two reasons are just based on my gut feeling. I'd like to understand this properly, and answering this for me would be awesome, so thanks in advance! I equally want to know what the flaws there are in my thinking :)
Thanks!

Homework Helper

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## The Attempt at a Solution

I know that given $f(x)=c$, the integral from $a≤x≤b$ is $c(b-a)$ (at least, I hope I know that! :D).
Is the integral the same value if you don't include an endpoint? That is, if you were evaluating $f(x)$ from $a≤x<b$?

Intuitively I think it both integrals should be the same value, for 2 reasons:

1) The two integrals are the same except one of them doesn't include the ordinate of b (does not include $f(b)$ in the set of function values). This can be represented as a vertical line which would have no area.

2) I'm even less sure about this bit, but surely you could use a limit to show that evaluating the integral from $a≤x<b$ - "from a to as close to be as you can get" - is the same as evaluating it up to and including b?

Like I said, I don't really know, and these two reasons are just based on my gut feeling. I'd like to understand this properly, and answering this for me would be awesome, so thanks in advance! I equally want to know what the flaws there are in my thinking :)
Thanks!

So your question is about the operator : $\int_{a}^{b}$

I believe what you're asking is what if we had : $\int_{a}^{}f(x)dx$

The answer is that this would not happen.

Flumpster
I think you misunderstood me, I probably haven't been using the correct terminology.
I dug up my calculus book, and I think what I'm thinking about is a step function.
In the book, it says the following:

"...if s(x) = c for all x in the closed interval [a,b], the ordinate set of s is a rectangle of base (b-a) and altitude c; the integral of s is c(b-a), the area of this rectangle. Changing the value of s at one or both endpoints a or b changes the ordinate set but does not alter the integral of s or the area of its ordinate set."

Lets say that f(x) = 5 for all x in the closed interval [a,b]. Or, put in a way more congruent with how I put it in my original post, x=5 when a ≤ x ≤ b. The integral is 5(b-a).

Now I change the value of the endpoint b to 7. x=5 when a ≤ x < b, and at b, x=7. Is the integral still given by 5(b-a)?

Homework Helper
I think you misunderstood me, I probably haven't been using the correct terminology.
I dug up my calculus book, and I think what I'm thinking about is a step function.
In the book, it says the following:

"...if s(x) = c for all x in the closed interval [a,b], the ordinate set of s is a rectangle of base (b-a) and altitude c; the integral of s is c(b-a), the area of this rectangle. Changing the value of s at one or both endpoints a or b changes the ordinate set but does not alter the integral of s or the area of its ordinate set."

Lets say that f(x) = 5 for all x in the closed interval [a,b]. Or, put in a way more congruent with how I put it in my original post, x=5 when a ≤ x ≤ b. The integral is 5(b-a).

Now I change the value of the endpoint b to 7. x=5 when a ≤ x < b, and at b, x=7. Is the integral still given by 5(b-a)?

I think I understand what you're trying to say, but think about it like this.

A more general way of thinking about what you're trying to say is to suppose f(x) = c ( Where c is a real number of course ). Visually you should know that f(x) = c is a straight, horizontal line which is parallel to the x-axis.

You also know that the limits of integration a and b are straight vertical lines which intersect f(x) and will allow you to calculate the area under f(x).

Now consider :

$\int_{a}^{b}f(x)dx = \int_{a}^{b}cdx = c\int_{a}^{b}dx = c[x]_{a}^{b} = c(b-a)$

Lets say c = 1, 2, 3, 4,...., n. The larger c becomes, the higher the height of the rectangle will be ( When I say height, I mean width in this case since we're working in 2D ). Changing the value of c will have absolutely no impact on the result of your integral other than a different number getting spat out of c(b-a). Changing b or a will also have no effect.

Yes, you can even change the value of your function at infinitely-many points:

Take the function that is 1 when x is irrational , and 0 when x is rational. Its

integral equals 1. The reason is, we can make the base rectangle as small as we

want, to make the contribution of the "bad" points to be negligible. More

formally, the integral is not affected by "sets of measure zero".

Flumpster
Thank you both for your answers :)
I still haven't really managed to explain what I'm talking about...I'll have a look at related threads in the forum to see if I can clarify.

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Homework Helper
Gold Member

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## The Attempt at a Solution

I know that given $f(x)=c$, the integral from $a≤x≤b$ is $c(b-a)$ (at least, I hope I know that! :D).
Is the integral the same value if you don't include an endpoint? That is, if you were evaluating $f(x)$ from $a≤x<b$?
(I know it's been answered.)

The answer is, yes, you get the same value if you don't include the end point.

You then have $\displaystyle \lim_{u\to b}\int_{a}^{u}f(x)\ dx=\lim_{u\to b}(c(u-a))=c(b-a)\ .$

Flumpster
Thanks everyone!

SammyS, thanks for writing it out like that, it made it a lot clearer to me.

RGV, I'll look at that link, thanks.

:)