Integration by Substitution/Primitive/Proof 1

Theorem

The primitive of $f$ can be evaluated by:

$\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $x = \map \phi u$.

Let $\map F x = \ds \int \map f x \rd x$.

Thus by definition $\map F x$ is a primitive of $\map f x$.

$\ds \map {\frac \d {\d u} } {\map F x}$

$=$

$\ds \map {\frac \d {\d u} } {\map F {\map \phi u} }$

Definition of $\map \phi u$

$\ds $

$=$

$\ds \dfrac \d {\d x} \map F {\map \phi u} \dfrac \d {\d u} \map \phi u$

$\ds $

$=$

$\ds \dfrac \d {\d x} \map F x \dfrac \d {\d u} \map \phi u$

Definition of $\map \phi u$

$\ds $

$=$

$\ds \map f x \dfrac \d {\d u} \map \phi u$

as $\map F x = \ds \int \map f x \rd x$

So $\map F x$ is an antiderivative of $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$.

Therefore:

$\ds \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

$=$

$\ds \map F x$

$\ds $

$=$

$\ds \int \map f x \rd x$

where $x = \map \phi u$.

$\blacksquare$