Question

Здравствуйте. Вычислите интегралы:​

Answer 1

Ответ:

$\displaystyle 2)\ \ \int\limits_0^{1/3}\,(3x+1)^3\, dx=\frac{(3x+1)^4}{3\cdot 4}\Big|_0^{1/3}=\frac{1}{12}\cdot (2^4-1^4)=\frac{15}{12}=\frac{5}{4}\\\\\\4)\ \ \int\limits^0_{-1}\, \frac{(1-x)^4}{7}dx=\frac{1}{7}\cdot \frac{(1-x)^5}{5}\Big|_{-1}^0=\frac{1}{35}\cdot (1^5-2^5)=-\frac{31}{35}$

$\displaystyle 2)\ \ \int\limits_{-8}^{-3}\frac{1}{\sqrt{1-x}}\, dx=-2\sqrt{1-x}\Big|_{-8}^{-3}=-2\cdot (\sqrt{4}-\sqrt{9})=-2\, (2-3)=2\\\\\\4)\ \ \int\limits^{47}_{14}\,\frac{4}{\sqrt{x+2}}\, dx=4\cdot 2\sqrt{x+2}\Big|_{14}^{47}=8\cdot (\sqrt{49}-\sqrt{16})=8\cdot (7-4)=24$