Primitive of Power

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Theorem

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.


That is:

$\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.


Proof

\(\ds \map {\frac \d {\d x} } {\dfrac {x^{n + 1} } {n + 1} }\) \(=\) \(\ds \paren {n + 1} \paren {\dfrac {x^{\paren {n + 1} - 1} } {n + 1} }\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds x^n\)
\(\ds \leadsto \ \ \) \(\ds \int x^n \rd x\) \(=\) \(\ds \frac {x^{n + 1} } {n + 1} + C\) Definition of Primitive (Calculus)

When $n = -1$ we have $n + 1 = 0$, and $\dfrac {x^{n + 1} } {n + 1} = \dfrac {x^0} 0$ is undefined.

$\blacksquare$


Also known as

Some sources refer to this as the reverse power rule, as it is the "reverse" of the Power Rule for Derivatives.

It is even suggested that it could be called the anti-power rule, but this appears to be unlikely to catch on.


Examples

Primitive of $x$ with respect to $x$

$\ds \int x \rd x = \dfrac {x^2} 2 + C$


Primitive of $\sqrt x$ with respect to $x$

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


Also see


Sources

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