# Function properties

• Nov 15th 2008, 11:16 AM
Isktaine
Function properties
I understand that if a function has more than one antiderivative because when you intergrate it you put +c (which can be any number) but does this still apply for a function on a certain interval eg (-2,2). And does this also apply for continuous functions? How does the antiderivative of continuous functions vary from just functions?

Hope that makes sense, I'm just trying to get my head around functions have antiderivatives and how many. Every textbook I read doesnt seem very clear.

Thanks,

Isk
• Nov 15th 2008, 01:16 PM
Mathstud28
Quote:

Originally Posted by Isktaine
I understand that if a function has more than one antiderivative because when you intergrate it you put +c (which can be any number) but does this still apply for a function on a certain interval eg (-2,2). And does this also apply for continuous functions? How does the antiderivative of continuous functions vary from just functions?

Hope that makes sense, I'm just trying to get my head around functions have antiderivatives and how many. Every textbook I read doesnt seem very clear.

Thanks,

Isk

The formal definition of an indefinite integral giving the antiderivative is if $\int{f(x)dx}=F(x)$, then we would have that $F(x)\equiv\left\{g(x):g'(x)=f(x)\right\}$. In words this just basically means that the antiderivative of a function is ANY function that when you take the derivative of it you get the orginal function. That is the need for the +C. For consider $\int{2x}dx=x^2+C$. We do not know what C is but no matter what it is when I take the derivative of $x^2+C$ I get $2x=f(x)$. To illustrate that if you asserted that $\int{2xdx}=x^2+1$ and only $x^2+1$ I would state that is one of them since $\frac{d}{dx}\bigg[x^2+1\bigg]=2x=f(x)$. But I would then say what about $x^2+2,x^2+5,x^2+10^{10}$? Because we have that $\frac{d}{dx}\bigg[x^2+2\bigg]=\frac{d}{dx}\bigg[x^2+5\bigg]=\frac{d}{dx}\bigg[x^2+10^{10}\bigg]=2x=f(x)$

The difference between a definite integral an indefinite integral is very simple. An indefinite as was shown above is a function, more accurately a set of functions, but a function nonetheless. But a definite integral is just a number.

To see this we need to accept the First Fundamental Theorem of Calculus which states that $\int_a^{b}f(x)dx=F(b)-F(a)~\text{where }F'(x)=f(x)$. You may interpret this as evaluating the antiderivative of the function at the upper bound and substracting the antiderivative at the lower bound. But you might ask what about the C? Well the gets nicely taken care of. Lets not just state the theorem but lets actually play it out a little more. Assume for a second that $\int_a^{b}f(x)dx=\bigg[\int{f(x)dx}\bigg]\bigg|_{x=a}^{x=b}$. Well we know that $\int{f(x)dx}=F(x)+C$. So now we rewrite our definite integral as follows

\begin{aligned}\int_a^{b}f(x)dx&=\bigg[\int{f(x)dx}\bigg]\bigg|_{x=a}^{x=b}\\
&=\bigg[F(x)+C\bigg]\bigg|_{x=a}^{x=b}\\
&=\bigg[F(b)+C\bigg]-\bigg[F(a)+C\bigg]\\
&=F(b)-F(a)+C-C\\
&=F(b)-F(a)
\end{aligned}

And since we know that when we evaluate a function at a number we just get a number.

Does that help at all?
• Nov 16th 2008, 01:08 AM
Isktaine
Yeah the FT of C makes more sense now!

Does that mean that a function on (-1,1) does have more than one antiderivative(regardless if it is continuous or not), ...but that it is not possible that a continuous function has no antiderivative? ..but a (non-continuous) function can have no antiderivative?

Since every continuous function has an antiderivative, is there a simple way to proof that?
• Nov 16th 2008, 02:45 AM
HallsofIvy
Quote:

Originally Posted by Isktaine
I understand that if a function has more than one antiderivative because when you intergrate it you put +c (which can be any number) but does this still apply for a function on a certain interval eg (-2,2).

The way you have asked that question, the answer is "yes". If a function, f, is defined, say, on [-2, 2], there still exist an infinite number of functions having f as derivative.

However, I suspect you really mean to ask about integrating a function on a an interval, say from -2 to 2. If f(x) is defined on [-2, 2], then there exist an infinite number of functions having f as derivative. If F(x) is one of those then any of them can be written F(x)+ c for some constant c. In that case, the integral of f, from -2 to 2, if (F(2)+ c)- (F(-2)+ c)= F(2)- F(-2). The "c" term cancels out and we get one specific number no matter which anti-derivative we use.

Quote:

And does this also apply for continuous functions?
Yes, of course. It applies to any integrable functions and all continuous functions are integrable.

Quote:

How does the antiderivative of continuous functions vary from just functions?
In no really important way. Anti-derivatives of functions that are NOT continuous at some point will not have a derivative at that point- but will still be continuous.

Quote:

Hope that makes sense, I'm just trying to get my head around functions have antiderivatives and how many. Every textbook I read doesnt seem very clear.

Thanks,
Isk
All continous functions HAVE anti-derivatives and any bounded functions with only "jump" discontinuities at a finite number of points have anti-derivatives.

However, again you may be asking the wrong question. If a function HAS an anti-derivative, it does not follow that the anti-derivative is "easy" or in fact any function you have ever seen before. In a very precise sense, 'almost all' anti-derivatives are NOT "elementary functions" (functions made from polynomials, rational functions, exponentials, logarithms, etc.)

For example, $f(x)= e^{-x^2}$ is continous and so has an anti-derivative. That anti-derivative is, in fact, very important in applications, especially to probabilty and statistics. But that anti-derivative cannot be written in terms of "elementary functions". The anti-derivative of $e^{-x^2}$ is (proportional to) "Erf(x)", a function use a lot in probability and statistics which is DEFINED as the antiderivative of (a multiple of) $e^{-x^2}$.
• Nov 16th 2008, 02:48 AM
HallsofIvy
[QUOTE=Isktaine;221444]Yeah the FT of C makes more sense now!

Does that mean that a function on (-1,1) does have more than one antiderivative(regardless if it is continuous or not), ...but that it is not possible that a continuous function has no antiderivative? ..but a (non-continuous) function can have no antiderivative?[quote]
Yes.

Quote:

Since every continuous function has an antiderivative, is there a simple way to proof that?
Depends on what YOU mean by "simple". Every really good Calculus book will have a proof and any introductory "Mathematical Analysis" book.