Integral is a function ?!

shahar

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Is Integral a function?
Why yes? Why not?
 

MarkFL

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Usually, an indefinite integral of a function is a one-parameter family of functions.
 

Dr.Peterson

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Is Integral a function?
Why yes? Why not?
Please clarify what your question means. Are you asking about the definite integral of a function being a function (of the limits of integration?), or about an indefinite integral (an antiderivative) being a function, or perhaps about integration being an operator that takes a function to a new function (indefinite) or to a number (definite), or something else entirely?
 

shahar

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What is one-parameter function?
 

pka

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Please clarify what your question means. Are you asking about the definite integral of a function being a function (of the limits of integration?), or about an indefinite integral (an antiderivative) being a function, or perhaps about integration being an operator that takes a function to a new function (indefinite) or to a number (definite), or something else entirely?
Is Integral a function? Why yes? Why not?
To shahar: as put this is a yes or no question. As such we need for you to supply an answer to Prof. Peterson questions above\(\displaystyle ~\uparrow\)
I will say that historically the word integral meant number. It was a measure of things like area.
Now Leonard Gillman defined for \(\displaystyle x > 0,\quad L(x) = \int_1^x {\frac{1}{t}dt}\) for the so-called natural logarithm.
We look forward to your reply.
 

MarkFL

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What is one-parameter function?
Consider the function:

\(\displaystyle f(x)=2x\)

Hence:

\(\displaystyle \int f(x)\,dx=x^2+C\)

This (on the RHS) is a one-parameter family of functions. The constant of integration is the parameter, and this anti-derivative is a family of functions where the parameter \(C\) can be any real number.
 

apple2357

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I have always seen indefinite integration as an operator rather than a function??? Because it has the job of transforming from one space to another? But i could be making this up?
 

JeffM

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\(\displaystyle f(x,\ y) = x + y.\)

Is the operator a function, or is the function an operator?
 
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