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Question Number 54073 by cesar.marval.larez@gmail.com last updated on 28/Jan/19

Answered by estudiante last updated on 28/Jan/19

$$VemosqesunaintegralimpropiadetipoI:$$$$\underset{R\to \mathrm{\infty }}{\lim }{\int }_a^R{x}^{n}dx=\underset{R\to \mathrm{\infty }}{\lim }\mid \frac{x^{n+1}}{n+1}{\mid }_a^R=\underset{R\to \mathrm{\infty }}{\lim }(\frac{R^{n+1}}{n+1}-\frac{a^{n+1}}{n+1})=\mathrm{\infty }$$$$Laintegralporlotantodivergea+\mathrm{\infty }$$

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