Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 395

\[1)\ y = \frac{1}{x};y = 4x;\ x = 1;\text{\ y} = 0:\]

\[S = \int_{\frac{1}{2}}^{1}{\frac{1}{x}\text{dx}} + \int_{0}^{\frac{1}{2}}{4x\ dx} =\]

\[= \left. \ \ln|x| \right|_{\frac{1}{2}}^{1} + \left. \ 2x^{2} \right|_{0}^{\frac{1}{2}} =\]

\[= \ln 1 - \ln\frac{1}{2} + 2 \bullet \left( \frac{1}{2} \right)^{2} - 2 \bullet 0^{2} =\]

\[= \frac{1}{2} - \ln\frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2} - \ln\frac{1}{2}.\]

\[2)\ y = \frac{1}{x^{2}};\ y = x;\ x = 2;\text{\ y} = 0:\]

\[S = \int_{1}^{2}{\frac{1}{x^{2}}\text{dx}} + \int_{0}^{1}\text{x\ dx} =\]

\[= \left. \ \frac{x^{- 1}}{- 1} \right|_{1}^{2} + \left. \ \frac{x^{2}}{2} \right|_{0}^{1} =\]

\[= \frac{2^{- 1}}{- 1} - \frac{1^{- 1}}{- 1} + \frac{1^{2}}{2} - \frac{0^{2}}{2} =\]

\[= - \frac{1}{2} + 1 + \frac{1}{2} = 1.\]

\[Ответ:\ \ 1.\]

\[3)\ y = x^{2} + 1;\ \ \ y = x + 1:\]

\[x^{2} + 1 = x + 1\]

\[x^{2} - x = 0\]

\[x(x - 1) = 0\]

\[x_{1} = 0;\text{\ \ \ }x_{2} = 1.\]

\[|S| = \int_{0}^{1}{\left( x^{2} + 1 - x - 1 \right)\text{dx}} =\]

\[= \int_{0}^{1}{\left( x^{2} - x \right)\text{dx}} = \left. \ \left( \frac{x^{3}}{3} - \frac{x^{2}}{2} \right) \right|_{0}^{1} =\]

\[= \left( \frac{1^{3}}{3} - \frac{1^{2}}{2} \right) - \left( \frac{0^{3}}{3} - \frac{0^{2}}{2} \right) =\]

\[= \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = - \frac{1}{6}.\]

\[Ответ:\ \ \frac{1}{6}.\]

\[4)\ y = x^{2} + 2;\ \ \ y = 2x + 2:\]

\[x^{2} + 2 = 2x + 2\]

\[x^{2} - 2x = 0\]

\[x(x - 2) = 0\]

\[x_{1} = 0;\text{\ \ \ }x_{2} = 2.\]

\[|S| = \int_{0}^{2}{\left( x^{2} + 2 - 2x - 2 \right)\text{dx}} =\]

\[= \int_{0}^{2}{\left( x^{2} - 2x \right)\text{dx}} = \left. \ \left( \frac{x^{3}}{3} - x^{2} \right) \right|_{0}^{2} =\]

\[= \left( \frac{2^{3}}{3} - 2^{2} \right) - \left( \frac{0^{3}}{3} - 0^{2} \right) =\]

\[= \frac{8}{3} - 4 = 2\frac{2}{3} - 4 = - 1\frac{1}{3}.\]

\[Ответ:\ \ 1\frac{1}{3}.\]