# Is there any difference at all between the antiderivative and the original function?

A definite integral on the interval $(a,b) =$ the antiderivative at $b -$ the antiderivative at $a$. Am I correct in saying that the antiderivative $=$ the original function? Because the integral is the derivative of the original function, and the answer to the integral is the original function... right?

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"Original function"...of what? The antiderivative of a function $\,f\,$, or better: a primitive function of $\,f\,$ is a function $\,F\,$ s.t. $\,F'=f\,$ over some open interval... – DonAntonio Jan 2 '13 at 4:46
Can we say, that the antiderivative is f(x)+C? So the antiderivative is the original function + any constant? – Richard Jan 2 '13 at 5:12
@Richard Posting the same comment 4 times is not very nice. I'm starting to doubt your humanity. Please show some thought if you're not a bot, these guys are genuinely trying to help you. – anegligibleperson Jan 2 '13 at 5:33
Not at all, @Richard: $$f(x)=\left(f(x)+C\right)=f'(x)\Longleftrightarrow f(x)'=f(x)=ke^x\,\,,\,k=\,\text{a constant}$$ so in most, MOST, cases, it is not true that $\,f(x) + C\,$ is the antiderivative of $\,f(x)\,$...It seems to be you must have misunderstood seriously something in this stuff, as you seem to have problems both with derivatives and indefinite integrals. Fortunately enough, it won't be hard, hopefully, for you to catch up, as this is very basic material and there exist thousands of books and sites where you can do it. – DonAntonio Jan 2 '13 at 12:07

Drive slow, homie.

The definite integral of $f(x)$ on $(a,b)$, written $\int_a^b f(x) \,dx$, is a number. This number is the (signed) area under the graph of $f(x)$ along $(a,b)$.

An antiderivative is a function $F(x)$ with the property that $F'(x) = f(x)$.

The Fundamental Theorem of Calculus tells us that we can compute definite integrals using antiderivatives, i.e. $\int_a^b f(x) \, dx = F(b) - F(a)$.

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Can we say, that the antiderivative is f(x)+C? So the antiderivative is the original function + any constant? – Richard Jan 2 '13 at 5:11

Am I correct in saying that the antiderivative = the original function?

Well, I can't tell what you are referring to by "original" function, but if you are asking if $\int f(x)\,dx = f(x)$, of course that is not true.

Just try one or two of the simplest examples (such as $\int x\,dx$ or $\int x^2\,dx$ or $\int \sin x\,dx$, etc.) you can think of and you will see this is not true.

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Can we say, that the antiderivative is f(x)+C? So the antiderivative is the original function + any constant? – Richard Jan 2 '13 at 5:11
Certainly not. $\int x\,dx= {x^2\over 2}+C$ and clearly $x\not={x^2\over 2}+C$ for any value of $C$. The same goes for all the other examples I listed. Besides, if $\int f(x)\,dx=f(x)+C$, then $\int$ isn't very interesting, is it? And why have the new term antiderivative to describe such a trivial action? – JohnD Jan 2 '13 at 15:08

So the fundamental theorem of calculus says that $$\int_a^b f'(x) dx = f(b) - f(a).$$

Can you think of any two functions which have the same derivative? If you could, then you would know that the antiderivative isn't necessarily the original function. Suppose $f$ and $g$ have the same derivative. Then by linearity, the function $f-g$ must have a derivative that is identically zero. What functions have derivatives which are identically zero?

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Can we say, that the antiderivative is f(x)+C? So the antiderivative is the original function + any constant? – Richard Jan 2 '13 at 5:09
Yup, that's exactly it! – Twiceler Jan 2 '13 at 5:51
Thanks so much! – Richard Jan 2 '13 at 5:57
@Twiceler, think twice before you answer positively such a question, as it seems to be Richard was asking you whether the antiderivative of a function $\,f(x)\,$ is $\,f(x)+C\,$ , which in general is as false as an intelligent and well-intentioned politician. – DonAntonio Jan 2 '13 at 12:10
Yes, to be clear, given a function $f$, the antiderivative of the derivative of $f$ is $f(x)+C$. Of course, the only functions for which $f'=f$ is $f(x)=ae^x$ (for any real $a$). – Twiceler Jan 2 '13 at 20:59