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Reference: G.H. Hardy - A Course of Pure Mathematics (ed. 3) page 229-230

I am working on the above book and am on the section where he introduces integrals. The way he begins to introduce them is to suppose there is a function $\psi(x)$. Then we wish to determine a function such that $\phi'(x)=\psi(x)$

He then provides his definition of the integral:

Definition: If $\psi(x)$ is the derivative of $\phi(x)$, then we call $\phi(x)$ an integral, or integral function of $\psi(x)$. The operation of forming $\psi(x)$ from $\phi(x)$ we call integration.

What I don't understand is why to go to the trouble of introducing $\psi(x)$. Instead of writing:

$\int\psi(x)dx$

it would be just as valid to write:

$\int\phi'(x)dx$

Without having to make the step of introducing $\psi(x)$.

Question: Why does Hardy introduce another new function instead of just using the derivative function?

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Because $\psi$ represents the function one is starting with; Hardy is saying: given some function $\psi$, then an "integral function of $\psi$" is defined to be any function $\phi$ such that $\phi^\prime = \psi$. –  kjo May 7 '13 at 15:06
    
@kjo your comment has cleared up my confusion. Could you rewrite your comment as an answer so that I may accept ti? –  GovEcon May 7 '13 at 19:42

1 Answer 1

up vote 4 down vote accepted

Hardy is saying that given some function $\psi$, then define an "integral function of $\psi$" to be any function $\phi$ such that $\phi^\prime = \psi$.

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Usually also called an "indefinite integral", or as I prefer, a "primitive". –  Pedro Tamaroff May 7 '13 at 22:14
    
And also "antiderivative", I know. I was just paraphrasing Hardy, as quoted by the OP. –  kjo May 7 '13 at 22:17

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