Using integration by parts, we find that `int x^(n)e^(-x) dx=`

Saturday, November 10, 2018

Using integration by parts, we find that $\displaystyle\int{{x}}^{{{n}}}{{e}}^{{-{x}}}{\left.{d}{x}\right.}=$

The integral
$\displaystyle\int{{x}}^{{n}}{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}$ has to be determined.

Integration by parts gives us the
rule: $\displaystyle\int{u}{d}{v}={u}\cdot{v}-\int{v}{d}{u}$

Let $\displaystyle{u}={{x}}^{{n}}$ and $\displaystyle{d}{v}={{e}}^{{-{{x}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{u}={n}\cdot{{x}}^{{{n}-{1}}}{\left.{d}{x}\right.}$

$\displaystyle{v}=-{{e}}^{{-{x}}}$

$\displaystyle\int{{x}}^{{n}}$
e^-x dx

$\displaystyle={{x}}^{{n}}\cdot-{1}\cdot{{e}}^{{-{{x}}}}-\int-{1}\cdot{{e}}^{{-{{x}}}}\cdot{n}\cdot{{x}}^{{{n}-{1}}}{\left.{d}{x}\right.}$

=
$\displaystyle{{x}}^{{n}}\cdot-{1}\cdot{{e}}^{{-{{x}}}}+{n}\cdot\int{{e}}^{{-{{x}}}}\cdot{{x}}^{{{n}-{1}}}{\left.{d}{x}\right.}$

= $\displaystyle-{{x}}^{{n}}\cdot{{e}}^{{-{{x}}}}+{n}\cdot\int{{e}}^{{-{{x}}}}\cdot{{x}}^{{{n}-{1}}}$
dx

$\displaystyle\int{{e}}^{{-{{x}}}}\cdot{{x}}^{{{n}-{1}}}{\left.{d}{x}\right.}$

= $\displaystyle-{{x}}^{{{n}-{1}}}\cdot{{e}}^{{-{{x}}}}+{\left({n}-{1}\right)}\cdot\int$
e^-x*x^(n-2) dx

Substituting this in the original integral

$\displaystyle\int{{x}}^{{n}}\cdot{{e}}^{{-{{x}}}}{\left.{d}{x}\right.}$

= $\displaystyle-{{x}}^{{n}}\cdot{{e}}^{{-{{x}}}}+{n}\cdot{\left(-{{x}}^{{{n}-{1}}}\cdot{{e}}^{{-{{x}}}}+{\left({n}-{1}\right)}\cdot\int\right.}$
e^-x*x^(n-2) dx)

= $\displaystyle-{{x}}^{{n}}\cdot{{e}}^{{-{{x}}}}-{n}\cdot{{x}}^{{{n}-{1}}}\cdot{{e}}^{{-{{x}}}}+{n}\cdot{\left({n}-{1}\right)}\cdot\int{{e}}^{{-{{x}}}}\cdot{{x}}^{{{n}-{2}}}$
dx

= $\displaystyle-{{e}}^{{-{{x}}}}\cdot{\left({{x}}^{{n}}+{n}\cdot{{x}}^{{{n}-{1}}}\right)}+{n}\cdot{\left({n}-{1}\right)}\cdot\int{{e}}^{{-{{x}}}}\cdot{{x}}^{{{n}-{2}}}{\left.{d}{x}\right.}$

This can be continued n times to yield the final result.

=
$\displaystyle-{{e}}^{{-{{x}}}}\cdot{\left({{x}}^{{n}}+{n}\cdot{{x}}^{{{n}-{1}}}+{n}\cdot{\left({n}-{1}\right)}{{x}}^{{{n}-{2}}}+\ldots{n}!\right)}$

The integral...

Spooky Notes

Saturday, November 10, 2018

Using integration by parts, we find that $\displaystyle\int{{x}}^{{{n}}}{{e}}^{{-{x}}}{\left.{d}{x}\right.}=$

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