Integral of Power/Conventional Proof

Theorem
$\displaystyle \forall n \in \R, n \ne -1: \int_0^b x^n \mathrm d x = \frac {b^{n+1}} {n+1}$

Proof
From the Fundamental Theorem of Calculus:


 * $(1): \quad \displaystyle \int_0^b x^n \ \mathrm d x = \left[{F \left({x}\right)}\right]_0^b = F \left({b}\right) - F \left({0}\right)$

where $F \left({x}\right)$ is a primitive of $x^n$.

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

Then: