Integration by Substitution/Primitive/Proof 1
Jump to navigation
Jump to search
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 $x = \map \phi u$.
Proof
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 $\map \phi u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac \d {\d x} \map F {\map \phi u} \dfrac \d {\d u} \map \phi u\) | Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac \d {\d x} \map F x \dfrac \d {\d u} \map \phi u\) | 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 $x = \map \phi u$.
$\blacksquare$