# Integral is a function ?!

#### shahar

##### Junior Member
Is Integral a function?
Why yes? Why not?

#### MarkFL

##### Super Moderator
Staff member
Usually, an indefinite integral of a function is a one-parameter family of functions.

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#### Dr.Peterson

##### Elite Member
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?

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#### shahar

##### Junior Member
What is one-parameter function?

#### pka

##### Elite Member
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.

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#### MarkFL

##### Super Moderator
Staff member
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.

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#### apple2357

##### Junior Member
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

##### Elite Member
$$\displaystyle f(x,\ y) = x + y.$$

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

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