# FTC Question

1. Aug 1, 2014

### 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,:

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

2. Aug 2, 2014

### 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.

3. Aug 2, 2014

### HallsofIvy

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

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).

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, $C= F_1(a)- F_2(a)$.)

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.

4. Aug 2, 2014

### Dethrone

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

5. Aug 2, 2014

### Dethrone

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.

6. Aug 2, 2014

### 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.

7. Aug 2, 2014

### 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.

8. Aug 2, 2014

### 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?

9. Aug 2, 2014

### 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.

10. Aug 2, 2014

### 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?

Last edited: Aug 2, 2014
11. Aug 2, 2014

### 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?

12. Aug 3, 2014

### HallsofIvy

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

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 $F(b)= \int_a^b f(t)dt$.

13. Aug 3, 2014

### 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.

14. Aug 3, 2014

### 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.

15. Aug 3, 2014

### 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?

Last edited: Aug 3, 2014
16. Aug 3, 2014

### HallsofIvy

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

17. Aug 3, 2014

### Dethrone

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