So, We all know that the definition of the derivative is
$f'(x)=\lim_{h\to0}{f(x+h)-f(x)\over h}$
So does the antiderivative have a solid definition like this?
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By 'solid definition' do you mean defined as a limit? – Olivier Oloa Oct 24 '16 at 07:56
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Of course it does. – StubbornAtom Oct 24 '16 at 08:14
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The antiderivative of a function $f$ on an interval $(a,b)$ is defined as the set of all functions $g$ such that $g'=f$ identically on $(a,b)$.
This definition is solid, despite your opinion. Since there is no unique antiderivative, there can be no formula whose result must be a single function. This is related to the fact that the operator $f \mapsto f'$ is not really invertible (on suitable spaces), since all constants have null derivative. To be consistent with this, we can only define the antiderivative as a set of functions.
Siminore
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Unfortunately, the antiderivative is the function F(x) such that F'(x)=f(x)...
Which explains why it is harder to integrate than to differentiate (just like for division VS multiplication)
Edouard L.
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