Primitive of Power of x by Exponential of a x/Examples/x cubed by e^-x
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Example of Use of Primitive of $x^n e^{a x}$
- $\ds \int x^3 e^{-x} \rd x = -e^{-x} \paren {x^3 + 3 x^2 + 6 x + 6} + C$
Proof
From Primitive of $x^n e^{a x}$:
Let $n$ be a positive integer.
Let $a$ be a non-zero real number.
Then:
\(\ds \int x^n e^{a x} \rd x\) | \(=\) | \(\ds \frac {e^{a x} } a \paren {x^n - \dfrac {n x^{n - 1} } a + \dfrac {n \paren {n - 1} x^{n - 2} } {a^2} - \dfrac {n \paren {n - 1} \paren {n - 2} x^{n - 3} } {a^3} + \cdots + \dfrac {\paren {-1}^n n!} {a^n} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C\) |
where $n^{\underline k}$ denotes the $k$th falling factorial power of $n$.
Hence:
\(\ds \int x^3 e^{-x} \rd x\) | \(=\) | \(\ds \frac {e^{-x} } {-1} \paren {x^3 - \dfrac {3 x^{3 - 1} } {-1} + \dfrac {3 \paren {3 - 1} x^{3 - 2} } {\paren {-1}^2} - \dfrac {3 \paren {3 - 1} \paren {3 - 2} x^{3 - 3} } {\paren {-1}^3} } + C\) | setting $a \gets -1$, $n \gets 3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -e^{-x} \paren {x^3 + 3 x^2 + 6 x + 6} + C\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $26$.