The domain of the derivative function, denoted as f'(x), is always less than or equal to the domain of the original function f(x). This principle holds true except in artificial cases where a function g has a larger domain than f and is equal to f' on the domain of f. Additionally, while the integral of f(x), represented as F(x), may have a domain equal to or greater than f(x), it does not always represent the area under the graph due to potential singularities and the nature of the function itself. For example, the integral of sin(x) from 0 to 2π equals zero, despite the presence of area beneath the curve.
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Thank you, if you don't mind I'd like to extrapolate off this idea a little.fresh_42 said:Yes. With the exception that if you consider two functions ##f,g## where ##g## has a different (bigger) domain than ##f## and it happens that ##g=f'## on the domain of ##f##. A bit artificial, I know, but the question seems a bit strange, too. That's because the first derivative of a function is a linear approximation of the function and can thus only be defined where the function itself is defined. Maybe not at some points like singularities or boundaries and therefore smaller.
Not really. As you define the derivative as well as the integral from the perspective of ##f##, how should they be defined elsewhere than ##f\,##? In the case of integrals, one might integrate even at points where ##f## isn't defined, e.g. at removable singularities (holes in the graph of ##f\,##). In addition there exist different concepts of integration (cp. https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/). But in principle you are stuck with ##f## as it is the basis of your considerations.Saracen Rue said:Thank you, if you don't mind I'd like to extrapolate off this idea a little.
I'm going to go ahead and assume that because the domain of ##f'(x)## is less than or equal to the domain of ##f##, the integral of ##f(x)##, ##F(x)## will have a domain equal to or greater than ##f(x)##.
Usually it is defined that way: ##f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}## or ##f(x+v)=f(x) + \nabla_x(f)\cdot v + r(v)## with a small correction ##r##.Now for another little question: while I do know for sure that ##f'(x)## is always the gradient of ##f(x)## at any value of ##x##, ...
Not really. The easy answer is no, because ##\int_0^{2\pi}\sin(x)\,dx=0## whereas there is actually an area beneath the sine. Or take the example with a removable singularity, where area isn't really defined. In addition it depends on ##f## itself. The picture with the area is a rather real one, i.e. for real functions in one variable. For complex functions this picture breaks down.... does it always also hold true that ##\int _a^bf\left(x\right)dx## will always give you the area between ##f(x)##, the ##x##-axis, and the lines ##x=a## and ##x=b##?
Does my example with the sine count? I don't know a good answer actually, as there are really pathological functions out there and the risk is high to forget some of those. Despite this we probably would have to narrow down the variety of possibilities: which functions of how many variables where from defined where with or without singularities and in the end eventually with which measure. For continuous real functions ##f\, : \,\mathbb{R} \rightarrow \mathbb{R}## between two consecutive zeros, the absolute value of the integral of ##f## is the area.Or is it possible for an integral to have a value that does not reflect the area under the graph?