Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1538

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

\[\sqrt{x - 1} = 3 - x\]

\[x - 1 = 9 - 6x + x^{2}\]

\[x^{2} - 7x + 10 = 0\]

\[D = 49 - 40 = 9\]

\[x_{1} = \frac{7 - 3}{2} = 2;\]

\[x_{2} = \frac{7 + 3}{2} = 5.\]

\[Имеет\ решения\ при:\]

\[3 - x \geq 0\]

\[x \leq 3.\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

\[x = 2.\]

\[\sqrt{x - 1} > 0\]

\[x - 1 \neq 0\]

\[x \neq 1.\]

\[3 - x > 0\]

\[x < 3.\]

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

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

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

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

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

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

\[Ответ:\ \ 1\frac{1}{6}.\]

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

\[- \frac{1}{x} = x^{2}\ \ \ \ \ | \bullet x;\]

\[- 1 = x^{3}\]

\[x = - 1.\]

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

\[- 8 = x^{3}\]

\[x = - 2.\]

\[x^{2} = \frac{x^{2}}{8}\ \ \ \ \ | \bullet 8\]

\[8x^{2} = x^{2}\]

\[7x^{2} = 0\]

\[x = 0.\]

\[S = \int_{- 2}^{- 1}\left( - \frac{1}{x} - \frac{x^{2}}{8} \right) + \int_{- 1}^{0}\left( x^{2} - \frac{x^{2}}{8} \right) =\]

\[= \int_{- 2}^{- 1}\left( - \frac{1}{x} - \frac{x^{2}}{8} \right) + \int_{- 1}^{0}\left( \frac{7}{8}x^{2} \right) =\]

\[= \left. \ \left( - \ln x - \frac{1}{8} \bullet \frac{x^{3}}{3} \right) \right|_{- 2}^{- 1} + \left. \ \left( \frac{7}{8} \bullet \frac{x^{3}}{3} \right) \right|_{- 1}^{0} =\]

\[= \left. \ \left( - \ln x - \frac{x^{3}}{24} \right) \right|_{- 2}^{- 1} + \left. \ \left( \frac{7x^{3}}{24} \right) \right|_{- 1}^{0} =\]

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

\[Ответ:\ \ \ln 2.\]