# Math Help - Quick puzzling question (antiderivatives)

1. ## Quick puzzling question (antiderivatives)

I am having some trouble with antiderivatives. On occasion I see something like:

The antiderivative of f(x) $= \int_{0}^{x} f(x')dx'$

I don't understand why this is used since it isn't true for something as simple as f(x) = e^x, in which case it is off by a constant. Can someone let me know where my understanding has gone wrong and why this notation is used?

2. Originally Posted by Mentia
I am having some trouble with antiderivatives. On occasion I see something like:

The antiderivative of f(x) $= \int_{0}^{x} f(x')dx'$

I don't understand why this is used since it isn't true for something as simple as f(x) = e^x, in which case it is off by a constant. Can someone let me know where my understanding has gone wrong and why this notation is used?
I have never seen it written that way. All an antiderivative does is "undo" the derivative. The way this is written is redundant because when f(x) is written as an integral it is already the derivative. So, the antiderivative is usually written as F(x) to show that it was originally in the form of a derivative.

3. Okay I think I understand what's going on in case anyone was wondering (probably not). It is helpful notation when you don't know what f(x) actually is and you are just writing the general case. For instance:

$\frac{d }{dx } g(x) = f(x)$

Then,
$
g(x) = \int_{ 0}^{ x} f(x')dx' + C1$

Notice this is slightly different from

$
g(x) = \int f(x)dx$

$g(x) = F(x) + C2$

where F(x) is the antiderivative of f(x). It should still be true in general. Notice it makes things slightly easier when you have boundary conditions at zero. For instance, if your condition is g(0) = 1 this immediately gives C1 = 1, whereas you have no knowledge about C2 without knowing f(x).

Similarly if your conditions are at some other constant, K, you can integrate from K to x, and then your constant of integration is equal to whatever your boundary condition value is:

$g(x) = \int_{ K}^{ x} f(x')dx' + C1$, $g(K)=Q$ --> $C1 = Q$