Question Number 54073 by cesar.marval.larez@gmail.com last updated on 28/Jan/19
Answered by estudiante last updated on 28/Jan/19
$${Vemos}\:{q}\:{es}\:{una}\:{integral}\:{impropia}\:{de}\:{tipo}\:{I}: \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\int_{{a}} ^{{R}} {x}^{{n}} {dx}\:=\:\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{{a}} ^{{R}} =\:\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{R}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:−\frac{{a}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)=\:\infty \\ $$$${La}\:{integral}\:{por}\:{lo}\:{tanto}\:{diverge}\:{a}\:+\infty \\ $$