# Integral of Power/Conventional Proof

## Theorem

$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

## Proof

From the Fundamental Theorem of Calculus:

$(1): \quad \ds \int_0^b x^n \rd x = \bigintlimits {\map F x} 0 b = \map F b - \map F 0$

where $\map F x$ is a primitive of $x^n$.

By Primitive of Power, $\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.

Then:

 $\ds \int_0^b x^n \rd x$ $=$ $\ds \intlimits {\frac {x^{n + 1} } {n + 1} } 0 b$ substituting $\dfrac {x^{n + 1} } {n + 1}$ for $F$ in $(1)$ $\ds$ $=$ $\ds \frac {b^{n + 1} } {n + 1} - \frac {0^{n + 1} } {n + 1}$ $\ds$ $=$ $\ds \frac {b^{n + 1} } {n + 1}$

$\blacksquare$