Is the anti-derivative another way of saying the derivative?

Question

This question is being posted so that I can understand a definition and concept.

The definition - F(x) is an anti-derivative of f(x) if F'(x) = f(x)

Therefore, $$F(x) = x^2$$ is an anti-derivative of $$f(x) = 2x$$ since $$F'(x) = 2x = f(x)$$

Answer 1

Your title sounds like (and generally is) a contradiction. However, your definition of anti-derivative is good. The relationship is $$ F \textrm{ is an antiderivative of } f \iff f \textrm{ is the derivative of } F $$

Answer 2

It is true that "$F$ is an antiderivative of $f$" is just another way of saying "$f$ is the derivative of $F$". These two sentences mean exactly the same thing: they both mean "$F' = f$". But take note of the terminology an antiderivative versus the derivative. We use this terminology because:

• If the derivative of a function $f$ exists, i.e. if $f'$ exists, then it is unique. There is only one derivative of $f$.

• But, if $f$ has an antiderivative, say $F$, then it is not the only one. In fact, $F + C$ is an antiderivative of $f$ for any constant $C$. So there are many antiderivatives of $f$, and $F$ is just one of them.

It is also worth noting:

• If $F$ is an antiderivative of $f$, then we sometimes say instead it is an indefinite integral of $f$, and write $F = \int f$.

• However, some functions are integrable but don't have an antiderivative. For instance, $f(x) = \begin{cases} 1 & \text{if } x = 0 \\ 0 & \text{otherwise} \end{cases}$ is integrable with integral $F(x) = 0$. But $F$ is not an antiderivative of $f$ since $F'$ is not equal to $f$.

So, roughly speaking, antiderivative is another term for indefinite integral, but not quite.

Answer 3

You know how to take a derivative. When you find an antiderivative, you are running the process in reverse.

Since the derivative of $x^2$ is $2x$, an antiderivative of $2x$ is $x^2$.

For more information, see this article.

Answer 4

Okay, this is a decent question as the terminology can get confusing so lets be very clear and precise about this: $$x^2+k$$ is the 'indefinite integral' of $$2x$$ since $$x^2+k$$ represents a "family" of quadratic curves all differing by a constant $k$. If we knew the value of $k$ we would then know the 'definite' integral.

$2x$ has many anti-derivatives (infinitely many); for example $x^2$ is an anti-derivative of $2x$ when $k=0$ and $x^2+4$ is another anti-derivative of $2x$ for the case when $k=4$ and so on.

Therefore an anti-derivative is just a particular case of the indefinite integral where you know the constant $k$; and hence an anti-derivative means exactly the same thing as a definite integral. Just two different words with the same meaning.

So the answer to the question in the title: "Is the anti-derivative another way of saying the derivative?" is no. But the rest you mentioned is fine. Just remember that indefinite integrals need a constant.