ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1096

\[1)\ y = \sqrt{x - 1};\ y = 3 - x;\ y = 0;\]

\[S = \int_{1}^{2}{(x - 1)^{\frac{1}{2}}\text{\ dx}} + \int_{2}^{3}{(3 - x)\text{dx}} =\]

\[= \left. \ \left( (x - 1)^{\frac{3}{2}}\ :\frac{3}{2} \right) \right|_{1}^{2} + \left. \ \left( 3x - \frac{x^{2}}{2} \right) \right|_{2}^{3} =\]

\[= \left. \ \left( \frac{2\sqrt{(x - 1)^{3}}}{3} \right) \right|_{1}^{2} + \left. \ \left( 3x - \frac{x^{2}}{2} \right) \right|_{2}^{3} =\]

\[S = \frac{2}{3} - \frac{0}{3} + 9 - \frac{9}{2} - 6 + \frac{4}{2} =\]

\[= \frac{2}{3} - \frac{5}{2} + 3 = \frac{4 - 15 + 18}{6} =\]

\[= \frac{7}{6} = 1\frac{1}{6}.\]

\[Ответ:\ \ 1\frac{1}{6}.\]

\[2)\ y = - \frac{1}{x};\ \ y = x^{2};\ \ \ y = \frac{x^{2}}{8};\]

\[= \ln\frac{| - 2|}{| - 1|} + \frac{1}{24} - \frac{8}{24} + \frac{0}{24} + \frac{7}{24} =\]

\[= \ln(2\ :1) = \ln 2.\]

\[Ответ:\ \ \ln 2.\]