ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 366

\[x = b;\ \ y = f(x).\]

\[1)\ b = 3;\text{\ \ \ f}(x) = x^{2}:\]

\[x^{3} = 0\]

\[x = 0.\]

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

\[= 3^{2} = 9.\]

\[Ответ:\ \ 9.\]

\[2)\ b = 2;\text{\ \ \ f}(x) = x^{3}:\]

\[x^{3} = 0\]

\[x = 0.\]

\[{S = \int_{0}^{2}{x^{3}\text{\ dx}} = \left. \ \frac{x^{4}}{4} \right|_{0}^{2} = \frac{2^{4}}{4} = }{= \frac{16}{4} = 4.}\]

\[Ответ:\ \ 4.\]

\[3)\ b = 4;\text{\ \ \ f}(x) = \sqrt{x}:\]

\[\sqrt{x} = 0\]

\[x = 0.\]

\[S = \int_{0}^{4}{x^{\frac{1}{2}}\text{\ dx}} = \left. \ \frac{2}{3}x^{\frac{3}{2}} \right|_{0}^{4} = \frac{2}{3} \bullet 4^{\frac{3}{2}} =\]

\[= \frac{2}{3} \bullet 2^{3} = \frac{16}{3} = 5\frac{1}{3}.\]

\[Ответ:\ \ 5\frac{1}{3}.\]

\[4)\ b = 8;\text{\ \ \ f}(x) = \sqrt[3]{x}:\]

\[\sqrt[3]{x} = 0\]

\[x = 0.\]

\[S = \int_{0}^{8}{x^{\frac{1}{3}}\text{\ dx}} = \left. \ \frac{3}{4}x^{\frac{4}{3}} \right|_{0}^{8} = \frac{3}{4} \bullet 8^{\frac{4}{3}} =\]

\[= \frac{3}{4} \bullet 2^{4} = \frac{3 \bullet 16}{4} = 12.\]

\[Ответ:\ \ 12.\]

\[5)\ b = 2;\text{\ \ \ f}(x) = 5x - x^{2}:\]

\[5x - x^{2} = 0\]

\[x(5 - x) = 0\]

\[x_{1} = 0;\text{\ \ \ }x_{2} = 5.\]

\[S_{1} = \int_{0}^{2}{\left( 5x - x^{2} \right)\text{dx}} =\]

\[= \left. \ \left( \frac{5x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{0}^{2} = \frac{5 \bullet 2^{2}}{2} - \frac{2^{3}}{3} =\]

\[= 10 - \frac{8}{3} = 7\frac{1}{3};\]

\[S_{2} = \int_{2}^{5}{\left( 5x - x^{2} \right)\text{dx}} =\]

\[= \left. \ \left( \frac{5x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{2}^{5} =\]

\[= \left( \frac{5 \bullet 5^{2}}{2} - \frac{5^{3}}{3} \right) - \left( \frac{5 \bullet 2^{2}}{2} - \frac{2^{3}}{3} \right) =\]

\[= \left( \frac{125}{2} - \frac{125}{3} \right) - \left( 10 - \frac{8}{3} \right) =\]

\[= \frac{375 - 250 - 60 + 16}{6} =\]

\[= \frac{81}{6} = 13\frac{1}{2}.\]

\[Ответ:\ \ 7\frac{1}{3};\ 13\frac{1}{2}.\]

\[6)\ b = 3;\text{\ \ \ f}(x) = x^{2} + 2x:\]

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

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

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

\[S = \int_{0}^{3}{\left( x^{2} + 2x \right)\text{dx}} =\]

\[= \left. \ \left( \frac{x^{3}}{3} + x^{2} \right) \right|_{0}^{3} = \frac{3^{3}}{3} + 3^{2} =\]

\[= 9 + 9 = 18.\]

\[Ответ:\ \ 18.\]

\[7)\ b = 1;\text{\ \ \ f}(x) = e^{x} - 1:\]

\[e^{x} - 1 = 0\]

\[e^{x} = 1\]

\[x = 0.\]

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

\[= \left( e^{1} - 1 \right) - \left( e^{0} - 0 \right) =\]

\[= (e - 1) - (1 - 0) = e - 2.\]

\[Ответ:\ \ e - 2.\]

\[8)\ b = 2;\text{\ \ \ f}(x) = 1 - \frac{1}{x}:\]

\[1 - \frac{1}{x} = 0\]

\[x = 1.\]

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

\[= \left. \ \left( x - \ln|x| \right) \right|_{1}^{2} =\]

\[= \left( 2 - \ln 2 \right) - \left( 1 - \ln 1 \right) =\]

\[= \left( 2 - \ln 2 \right) - (1 - 0) = 1 - \ln 2.\]

\[Ответ:\ \ 1 - \ln 2.\]