Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 397

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

\[y^{'} = 2x - 2 + 0 = 2x - 2;\]

\[y(0) = 0^{2} - 2 \bullet 0 + 2 = 2;\]

\[y^{'}(0) = 2 \bullet 0 - 2 = - 2;\]

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

\[|S| =\]

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

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

\[Ответ:\ \ \frac{1}{3}.\]

\[2)\ y = \frac{4}{x}:\]

\[y^{'} = 4 \bullet \left( - \frac{1}{x^{2}} \right) = - \frac{4}{x^{2}};\]

\[y(2) = \frac{4}{2} = 2;\]

\[y^{'}(2) = - \frac{4}{2^{2}} = - \frac{4}{4} = - 1;\]

\[y = 2 - 1 \bullet (x - 2) = 4 - x.\]

\[S = \int_{4}^{6}{\frac{4}{x}\text{dx}} + \int_{2}^{4}{\left( \frac{4}{x} - 4 + x \right)\text{dx}} =\]

\[= \left. \ 4\ln|x| \right|_{4}^{6} + \left. \ \left( 4\ln|x| - 4x + \frac{x^{2}}{2} \right) \right|_{2}^{4} =\]

\[= 4\left( \ln 6 - \ln 2 \right) - 16 + \frac{16}{2} + 8 - \frac{4}{2} =\]

\[= 4\ln 3 - 8 + 8 - 2 = 4\ln 3 - 2.\]

\[Ответ:\ \ 4\ln 3 - 2.\]