Does F(x) = F(x) - F(a)? Confused!

[Dethrone]

This is going to be a really dumb question, but somehow it's not really clicking.

I read somewhere that: \int_{a}^{x} f(t)\,dt = F(x) - F(a)

But, this is also true: F(x) = \int_{a}^{x} f(t)\,dt.

Does this imply that F(x) = F(x) - F(a)?

For example, at a random number, c,:

F(c) = \int_{c}^{x} f(t)\,dt = F(c) - F(a)

 

[WWGD]

A couple of problems, if you use the Riemann integral , F(x) with F'=f does not always exist. If it does, you use it in the top equation. But it is not the same F as the F in the second equation. And F(c) is the integral from a to c.

 

[HallsofIvy]

This is going to be a really dumb question, but somehow it's not really clicking.

I read somewhere that: \int_{a}^{x} f(t)\,dt = F(x) - F(a)

Yes, if F'(x)= f(x) this is true.

But, this is also true: F(x) = \int_{a}^{x} f(t)\,dt.

No, this is not true unless it happens that F(a)= 0. The difference is that F(x) is NOT 'well defined'. If F'= f then (F+ C)'= f for any constant C.

What is true is
1) \int_a^x f(t)dt= F(x)- F(a) for any function, F, such that F'= f
and
2) There exist a function, F, such that F'= f and \int_a^x f(t)dt= F(x)

The specific function, F, in (2) is one of the infinite number of functions in (1).

Does this imply that F(x) = F(x) - F(a)?

No. It implies that if F_1 and F_2 are any two of the infinite number of anti-derivatives of f, then F_1(x)= F_2(x)+ C for some constant, C. (Clearly, C= F_1(a)- F_2(a).)

For example, at a random number, c,:

F(c) = \int_{c}^{x} f(t)\,dt = F(c) - F(a)

No, the "F" on the left has to be a different anti-derivative of f than the "F" on the right. You are using the same notation to indicate two different functions.

 

[Dethrone]

Yes! I figured this out last night, as well! Thanks everyone!

 

[Dethrone]

Yes, if F'(x)= f(x) this is true.

2) There exist a function, F, such that F'= f and \int_a^x f(t)dt= F(x)

The specific function, F, in (2) is one of the infinite number of functions in (1).

Actually, I thought I figured it out, but I don't quite understand 2. Can you elaborate on that? I understand that there are infinite number of antiderivatives of a function, and F1(x) = F2(x) + c.

What I thought was that here F(x) = \int_{a}^{x} f(t)\,dt, I used F(x) to represent the function as a function of x, where as here \int_{a}^{x} f(t)\,dt = F(x) - F(a), F(x) is used to denote the anti derivative. Because they are used to denote different things, they are not equal. That's what I thought was the answer last night.

 

[HallsofIvy]

You are still using the phrase "the anti-derivative" though you say you understand "there are infinite number of antiderivatives of a function". When you write F(x)= \int_a^x f(t)dt you are saying that "F" is specifically that anti-derivative such that F(a)= 0.

 

[WWGD]

Besides, remember that even if the integral exists (when f is continuous outside of a set of measure zero), the integral may exist , i.e., the Riemann sums converge, but there may be no antiderivative F with F'=f. Notice that the contradiction in F(x)= \int_a^x f(t)dt=F(x)-F(a) shows the two functions cannot be equal, i.e., F cannot be used for both equations.

 

[Dethrone]

I realized that writing F(x)= \int_a^x f(t)dt implies the case when F(a) = 0, since \int_a^x f(t)dt = F(x) - F(a). I think I was confused because of notation: sometimes people write F(x) = \int_a^x f(t)dt to represent the integral as a function of x. i.e F(3) = \int_a^3 f(t)dt, and not using "F(x)" to refer to the antiderivative. When people write \int_a^x f(t)dt = F(x) - F(a), they are saying that the integral equals the antiderivative evaluated at x minus the antiderivative evaluated at a. So in this case, "F" is used to refer to the antiderivative. Am I right, or am I confused?

 

[HallsofIvy]

An anti-derivative, not "the" derivative! Other than that, yes, you are correct. Again, \int_a^x f(t)dt= F(x)- F(a) for F any anti-derivative of f.

 

[Dethrone]

Thought one: One thing, say we find the antiderivatives of y = x^2. Obviously, \int x^2 dx = \frac{x^3}{3} + C, where \frac{x^3}{3} + C represents all the possible antiderivatives. , i.e \frac{x^3}{3} + 4. But we won't know which one it is until we get a definite integral, right?

Thought two: Okay, if I'm getting it straight. Even if the definite integral is \int_{0}^{3} x^2\,dx = 9, there are still infinite antiderivatives of "F", right? For some odd reason, I used to think that the definite integral boils said infinite antiderivatives to just one.

Which "thought" is correct, or are they both wrong?

 

[Dethrone]

Wikipedia says (I think) that F(x) = \int_a^x f(t)dt produces different antiderivatives depending on which value of "x" you put in there. But last time I checked, definite integrals yield a value, which is the area under the curve on that interval. How does said "number" or "value" represent an antiderivative? Antiderivatives are suppose to be represented by a function, no?

 

[HallsofIvy]

Wikipedia says (I think) that F(x) = \int_a^x f(t)dt produces different antiderivatives depending on which value of "x" you put in there.

Are you sure it doesn't say "depending on which value of "a" you put in there?

But last time I checked, definite integrals yield a value, which is the area under the curve on that interval. How does said "number" or "value" represent an antiderivative? Antiderivatives are suppose to be represented by a function, no?

That is not a "definite integral" in the sense that you are using. The "x" is a variable and so F(x)= \int_a^x f(t)dt is a function of x not a number. For a specific value of x, say x= b, we have the number F(b)= \int_a^b f(t)dt.

 

[Dethrone]

But F(x)= \int_a^x f(t)dt doesn't represent antiderivatives right,? It just represents the area under the curve. If we put a value such as x = b like you said above: F(b)= \int_a^b f(t)dt this yields a number, not a function.

 

[HallsofIvy]

There is NO "area under the curve" unless you specify all boundaries of the region under the curve. \int_a^x f(t) dt doesn't "represent the area under the curve" because "x" is not a specific number so does not define a boundary. Again, F(x)= \int_a^x f(t) dt is one of the infinite number of antiderivatives of f.

 

[Dethrone]

Alright, so the other antiderivatives are related to F(x) = G(x) + C, where G(x) could be \int_b^x f(t) dt, by changing the value of "a" at the bottom? I was saying that F(c), where c is a number = F(c)= \int_c^b f(t)dt yields a number, because it has both boundaries.

EDIT: A moment of epiphany, I do hope this is right:

If we have F(x)= \int_a^x f(t) dt, we get all the different other antiderivatives if we change the value of "a"! Because \int_a^x f(t) dt = F(x) - F(a), where F(a) is a constant term, and we have \int_b^x f(t) dt = F(x) - F(b), where F(b) is a constant term. These all have the same derivative, namely, f(x) because the constant term disappears. Am I right now?

 

[HallsofIvy]

Yes, that's pretty much what I have been trying to say!

 

[Dethrone]

Alright, sorry if I been terribly slow. I finally understand all of this! Thanks HallsofIvy :D!