Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 382

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Год:2020-2021-2022-2023
Тип:учебник

Задание 382

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

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

\[x^{2} + 4x + 4 = x + 2\]

\[x^{2} + 3x + 2 = 0\]

\[D = 9 - 8 = 1\]

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

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

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

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

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

\[= \left( - \frac{1}{3} + \frac{3}{2} - 2 \right) - \left( - \frac{8}{3} + 6 - 4 \right) =\]

\[= \frac{7}{3} + \frac{3}{2} - 4 = \frac{14 + 9}{6} - 4 =\]

\[= \frac{23}{6} - 4 = 3\frac{5}{6} - 4 = - \frac{1}{6}.\]

\[Ответ:\ \ \frac{1}{6}.\]

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

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

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

\[|S| =\]

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

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

\[Ответ:\ \ \frac{1}{3}.\]

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

\[\sqrt{x} = x\]

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

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

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

\[= \left( \frac{2}{3} - \frac{1}{2} \right) - (0 - 0) = \frac{4 - 3}{6} = \frac{1}{6}.\]

\[Ответ:\ \ \frac{1}{6}.\]

\[4)\ y = (x - 1)^{2};\ y = 5 + x:\]

\[(x - 1)^{2} = 5 + x\]

\[x^{2} - 2x + 1 = 5 + x\]

\[x^{2} - 3x - 4 = 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[|S| = \int_{- 1}^{4}{\left( (x - 1)^{2} - (5 + x) \right)\text{dx}} =\]

\[= \int_{- 1}^{4}{\left( x^{2} - 3x - 4 \right)\text{dx}} =\]

\[= \left. \ \left( \frac{x^{3}}{3} - \frac{3}{2}x^{2} - 4x \right) \right|_{- 1}^{4} =\]

\[= \left( \frac{64}{3} - \frac{48}{2} - 16 \right) - \left( - \frac{1}{3} - \frac{3}{2} + 4 \right) =\]

\[= \frac{65}{3} - \frac{45}{2} - 20 = \frac{130 - 135}{6} - 20 =\]

\[= - \frac{5}{6} - 20 = - 20\frac{5}{6}.\]

\[Ответ:\ \ 20\frac{5}{6}.\]

\[5)\ y = \sin x;\ y = 1\ на\ \left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

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

\[= \left. \ \left( - \cos x - x \right) \right|_{0}^{\frac{\pi}{2}} =\]

\[= \left( - \cos\frac{\pi}{2} - \frac{\pi}{2} \right) - \left( - \cos 0 - 0 \right) =\]

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

\[Ответ:\ \ \frac{\pi}{2} - 1.\]

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