Primitive of Square Root

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Theorem

$\ds \int \sqrt x \rd x = \dfrac {2 x^{3 / 2} } 3 = \dfrac {2 \sqrt x^3} 3$


Proof

From Primitive of Power:

Let $n \in \R: n \ne -1$.


Then:

$\ds \int x^n \rd x = \frac {x^{n + 1} } {n + 1} + C$

where $C$ is an arbitrary constant.


Hence:

\(\ds \int \sqrt x \rd x\) \(=\) \(\ds \int x^{1/2} \rd x\) Definition of Square Root
\(\ds \) \(=\) \(\ds \dfrac {x^{1/2 + 1} } {1/2 + 1} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \dfrac {x^{3/2} } {3/2} + C\) simplification
\(\ds \) \(=\) \(\ds \dfrac {2 \sqrt x^3} 3 + C\) simplification

$\blacksquare$