Primitive of Root of Function under Half its Derivative
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Theorem
Let $f$ be a real function which is integrable.
Then:
- $\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x = \sqrt {\map f x} + C$
where $C$ is an arbitrary constant.
Proof
By Integration by Substitution (with appropriate renaming of variables):
- $\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$
Let $\map u x = \sqrt {\map f x}$
\(\ds \map u x\) | \(=\) | \(\ds \sqrt {\map f x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d u} {\d x}\) | \(=\) | \(\ds \dfrac {\map {f'} x} {2 \sqrt {\map f x} }\) | Chain Rule for Derivatives, Derivative of Power |
Then:
\(\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x\) | \(=\) | \(\ds \int \rd u\) | Integration by Substitution: putting $\map u x = \sqrt {\map f x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds u + C\) | Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\map f x} + C\) | Definition of $u$ |
$\blacksquare$
Sources
- 1960: Margaret M. Gow: A Course in Pure Mathematics ... (previous) ... (next): Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {IV}$.