Integral of Power/Conventional Proof

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Theorem

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


Proof

From the Fundamental Theorem of Calculus:

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

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:

\(\displaystyle \int_0^b x^n \mathrm d x\) \(=\) \(\displaystyle \left[{\frac {x^{n+1} } {n+1} }\right]_0^b\) substituting $\dfrac {x^{n+1} } {n+1}$ for $F$ in $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \frac {b^{n+1} } {n+1} - \frac {0^{n+1} } {n+1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {b^{n+1} } {n+1}\)

$\blacksquare$