Primitive of Power of x by Exponential of a x/Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n$ be a positive integer.

Then:

$\displaystyle \int x^n e^{a x} \rd x = \frac {x^n e^{a x} } a - \frac n a \int x^{n - 1} e^{a x} \rd x + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x^n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle n x^{n - 1}\) Derivative of Power


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle e^{a x}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {e^{a x} } a\) Primitive of Exponential of a x


Then:

\(\displaystyle \int x^n e^{a x} \rd x\) \(=\) \(\displaystyle x^n \frac {e^{a x} } a - \int \frac {e^{a x} } a \paren {n x^{n - 1} } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^n e^{a x} } a - \frac n a \int x^{n - 1} e^{a x} \rd x + C\) simplifying

$\blacksquare$


Sources