Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1540

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\[1)\ y = \sqrt{x};\ \ \ y = 2;\text{\ \ \ x} = 9:\]

\[\sqrt{x} = 2\]

\[x = 4.\]

\[S = \int_{4}^{9}\left( \sqrt{x} - 2 \right) = \int_{4}^{9}\left( x^{\frac{1}{2}} - 2 \right) =\]

\[= \left. \ \left( x^{\frac{3}{2}}\ :\frac{3}{2} - 2 \bullet \frac{x^{1}}{1} \right) \right|_{4}^{9} =\]

\[= \left. \ \left( \frac{2\sqrt{x^{3}}}{3} - 2x \right) \right|_{4}^{9} =\]

\[= \frac{2\sqrt{9^{3}}}{3} - 2 \bullet 9 - \frac{2\sqrt{4^{3}}}{3} + 2 \bullet 4 =\]

\[= \frac{2\sqrt{729}}{3} - 18 - \frac{2\sqrt{64}}{3} + 8 =\]

\[= \frac{2 \bullet 27}{3} - \frac{2 \bullet 8}{3} - 10 =\]

\[= \frac{54 - 16 - 30}{3} = \frac{8}{3} = 2\frac{2}{3}.\]

\[Ответ:\ \ 2\frac{2}{3}.\]

\[2)\ y = x^{2} + 3;\ \ \ y = x + 5:\]

\[x^{2} + 3 = x + 5\]

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

\[D = 1 + 8 = 9\]

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

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

\[S = \int_{- 1}^{2}\left( x^{2} + 3 - x - 5 \right) =\]

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

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

\[= \frac{2^{3}}{3} - \frac{2^{2}}{2} - 2 \bullet 2 - \frac{( - 1)^{3}}{3} + \frac{( - 1)^{2}}{2} + 2 \bullet ( - 1) =\]

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

\[= \frac{9}{3} - 2 - 6 + 0,5 =\]

\[= 3 - 7,5 = - 4,5.\]

\[Ответ:\ \ 4,5.\]