30 БАЛЛОВ!!! НОМЕР 169

Решите задачу:

$1)\int\limits^{-2} _{-3} {3x^{-2} } \, dx =-\frac{3}{x} |^{-2}_{-3}=-\frac{3}{-2}+\frac{3}{-3}=1,5-1=0,5\\\\2)\int\limits^{32} _{1}{\frac{1}{\sqrt[5]{x} } } \, dx= \frac{5\sqrt[5]{x^{4} } }{4}|_{1}^{32}=\frac{5}{4}(\sqrt[5]{32^{4} }-\sqrt[5]{1^{4} })=\frac{5}{4}(\sqrt[5]{(2^{4})^{5}}-1)=\frac{5}{4}(2^{4}-1)=\frac{5}{4}*15=\frac{75}{4}=18,75$ $3)\int\limits^{3} _{1}{(x^{3}+x)^{2}} \, dx =\int\limits^{3} _{1}{x^{6} +2x^{4} +x^{2} } \, dx=(\frac{x^{7} }{7}+\frac{2x^{5} }{5}+\frac{x^{3} }{3})|_{1}^{3}=\frac{3^{7} }{7}+\frac{2*3^{5} }{5}+\frac{3^{3} }{3} -\frac{1^{7} }{7}-\frac{2*1^{5} }{5}-\frac{1^{3} }{3}=\frac{2187}{7}+\frac{486}{5}+9-\frac{1}{7}-\frac{2}{5}-\frac{1}{3}=\frac{2186}{7}+\frac{484}{5}+9-\frac{1}{3}=\frac{43864}{105} =417\frac{79}{105}$