Question

определенный интеграл, помогите пожалуйста) ​

Answer 1

$1)\; \; \int\limits^8_0\, (\sqrt2+\sqrt[3]{x})\, dx=(\sqrt2\, x+\frac{3x^{4/3}}{4})\Big|_0^8=\sqrt2\cdot 8+12=4\, (2\sqrt2+3)\\\\\\2)\; \; \int\limits^4_1\, \frac{1+\sqrt{y}}{y^2}\, dy=(\frac{y^{-1}}{-1}+\frac{2y^{-1/2}}{-1})\Big|_1^4=(-\frac{1}{y}-\frac{2}{\sqrt{y}})\Big |_1^4=\\\\=-\frac{1}{4}-1+1+2=1\frac{3}{4}$

$3)\; \; \int\limits_0^{\pi /4}\, \frac{2\, dx}{cos^2x}\, dx=2\, tgx\Big |_0^{\pi /4}=2\cdot (1-0)=2\\\\\\4)\; \; \int\limits^2_1\, \frac{3x^4-5x^2+7}{x}\, dx=\int\limits^2_1(3x^3-5x+\frac{7}{x})\, dx=(\frac{3x^4}{4}-\frac{5x^2}{2}+7\, ln|x|)\Big|_1^2=\\\\=12-10+7\, ln2-(\frac{3}{4}-\frac{5}{2}+0)=0,25+7\, ln2$

$5)\; \; \int\limits^{2/3}_{-2/3}(x^3-2x)\, dx=(\frac{x^4}{4}-x^2)\Big|_{-2/3}^{2/3}=\frac{(2/3)^4}{4}-\frac{4}{9}-\frac{(2/3)^4}{4}+\frac{4}{9}=0$