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Show that [infinity] 0 e−x9 is convergent. SOLUTION We can't evaluate the integral directly because the antiderivative of e−x9 is not an elementary function. We write [infinity] 0 e−x9 dx = 1 0 e−x9 dx + [infinity] 1 dx and observe that the first integral on the right-hand side is just an ordinar

Respuesta :

Step-by-step explanation:

Just remember that

[tex]e^{-x^9} < e^{-x}[/tex]

That is the most important part, therefore,

[tex]\int\limits_{0}^{\infty}{e^{-x^9}} \, dx < \int\limits_{0}^{\infty}{e^{-x}} \, dx\\[/tex]

Since  

[tex]\int\limits_{0}^{\infty}{e^{-x}} \, dx = \lim\limits_{n \to \infty} -e^{-x} + e^{-1} = e^{-1}[/tex]

and the integral

[tex]\int\limits_{0}^{1} e^{-x^9} \, dx[/tex]

is an integral over a bounded interval and

[tex]\int\limits_{0}^{\infty} {e^{-x^9}} \, dx = \int\limits_{0}^{1} {e^{-x^9}} \, dx + \int\limits_{1}^{\infty} {e^{-x^9}} \, dx[/tex]

The integral

[tex]\int\limits_{0}^{\infty} {e^{-x^9}} \, dx[/tex]

converges.

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