Is f(x) an antiderivative of f'(x) or a family of antiderivatives?

[Jhenrique]

By FTC, every function f(x) can be expessed like: f(x) = \int_{x_0}^{x}f'(u)du + f(x_0) Now, I ask: f(x) is a antiderivative of f'(x) or is a family of antiderivative of f'(x) ?

 

[Mark44]

By FTC, every function f(x) can be expessed like: f(x) = \int_{x_0}^{x}f'(u)du + f(x_0) Now, I ask: f(x) is a antiderivative of f'(x) or is a family of antiderivative of f'(x) ?

f(x) is any antiderivative of f'.

 

[Jhenrique]

f(x) is any antiderivative of f'.

That's not what I asked.

 

[Mark44]

I actually did answer your question.

f(x) is a antiderivative of f'(x) or is a family of antiderivative of f'(x) ?

f is one (pick anyone that you like) antiderivative of f'. f is not a family of antiderivatives.

 

[Jhenrique]

Now yes!

So, you are saying (implicitly) that f(x₀) is not a arbitrary constant?

 

[Mark44]

Let's look at an example:
$$\int_1^x t^2 dt$$

Solution 1 -
One antiderivative of t² is (1/3)t³. Call this F₁(t). By the FTC, the integral above is equal to F₁(x) - F₁(1) = (1/3)x³ - 1/3

Solution 2
Another antiderivative is t² - 3. Call this F₂(t). Again, by the FTC, the integral above is equal to F₂(x) - F₂(1) = [(1/3)x³ - 3] - [1/3 - 3] = (1/3)x³ - 1/3. This is the same value that was obtained in solution 1.

What you get for F(t₀) depends on which function you use for the antiderivative, but the resulting value of the definite integral doesn't depend on which antiderivative you pick. You can choose any function in the family of antiderivatives, without making any difference at all in the resulting value of the definite integral. For this reason, it's most convenient to work with the antiderivative for which the constant is 0.

All of this seems very straightforward to me. Is there something I'm missing in what you're asking?

 

[Jhenrique]

Sorry my arrogance, but I still didn't understood if f(x₀) is a arbitrary value...

 

[Mark44]

It's arbitrary in the sense that f is an unspecified antiderivative. In my two examples f(1) has two different values, depending on which antiderivative we're using. In my first example, f(1) (which I'm calling F₁(1)) = 1/3. In the second example, F₂(1) = 1/3 - 3 = -8/3.

For a specified antiderivative F, F(t₀) is a constant, but if the particular antiderivative is not known, then we don't know the value of F(t₀).

 

[Jhenrique]

Based in your explanation, I can say that f(x₀) isn't an arbitrary constant such as is an arbritrary constant C, that appears in integration*. Correct!?

*∫ydx = Y + C

 

[Mark44]

Yeah, I think that's OK.