Интеграл................................................................

Решите задачу:

$\int _3^{+\infty}\frac{dx}{e^{\frac{x}{3}}}=lim_{b\to +\infty}\int _3^{b}\, e^{-\frac{x}{3}}\, dx=-3\cdot lim_{x\to +\infty }\int _3^{b}e^{-\frac{x}{3}}dx=\\\\=-3\cdot lim_{x\to +\infty }\int _3^{b}e^{-\frac{x}{3}}\, d(-\frac{x}{3})=-3\cdot lim_{x\to +\infty }e^{-\frac{x}{3}}\, |_3^{b}=\\\\=-3lim_{x\to +\infty }(e^{\frac{-b}{3}}-e^{-1})=-3\cdot (0-\frac{1}{e})=\frac{3}{e}=const\; \; \to \; \; sxoditsya$

$P.S.\; lim_{x\to \infty}e^{x}= \left \{ {{0,esli\; x\to -\infty } \atop {+\infty,esli\; x\to +\infty}} \right.$