Primitive of x by Exponential of a x

From ProofWiki
Jump to: navigation, search

Theorem

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


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:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

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


Then:

\(\displaystyle \int x e^{a x} \rd x\) \(=\) \(\displaystyle x \paren {\frac {e^{a x} } a} - \int \frac {e^{a x} } a \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle x \paren {\frac {e^{a x} } a} - \frac 1 a \int e^{a x} \rd x + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle x \paren {\frac {e^{a x} } a} - \frac 1 a \paren {\frac {e^{a x} } a} + C\) Primitive of $e^{a x}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{a x} } a \paren {x - \frac 1 a} + C\) simplifying

$\blacksquare$


Sources