Primitive of x squared by Exponential of x
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Theorem
- $\ds \int x^2 e^x \rd x = e^x \paren {x^2 - 2 x + 2} + C$
Proof 1
From Primitive of $x^2 e^{a x}$:
- $\ds \int x^2 e^{a x} \rd x = \frac {e^{a x} } a \paren {x^2 - \frac {2 x} a + \frac 2 {a^2} } + C$
The result follows by setting $a = 1$.
$\blacksquare$
Proof 2
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds 2 x\) | Derivative of Power |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds e^x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds e^x\) | Primitive of $e^x$ |
Then:
\(\ds \int x^2 e^x \rd x\) | \(=\) | \(\ds x^2 e^x - \int 2 x e^x \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 e^x - 2 \int x e^x \rd x + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 e^x - 2 \paren {e^x \paren {x - 1} } + C\) | Primitive of $x e^{a x}$ with $a = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e^x \paren {x^2 - 2 x + 2} + C\) | simplifying |
$\blacksquare$