Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1039

\[\boxed{\mathbf{1039}\mathbf{.}}\]

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

\[y = x^{2} + 4x + 4,\ \ \ y = 0\]

\[x^{2} - 6x + 9 = x^{2} + 4x + 4\]

\[4x + 6x = 9 - 4\]

\[10x = 5\ \]

\[x = \frac{1}{2}.\]

\[\textbf{а)}\ x^{2} - 6x + 9 > 0\]

\[(x - 3)^{2} > 0\]

\[x \neq 3.\]

\[\textbf{б)}\ x^{2} + 4x + 4 > 0\]

\[(x + 2)^{2} > 0\ \]

\[x \neq - 2.\]

\[= 11 - \frac{8}{2} + \frac{8}{3} + \frac{3}{4} =\]

\[= \frac{132 - 48 + 32 + 9}{12} = \frac{125}{12} =\]

\[= 10\frac{5}{12}.\]

\[Ответ:\ \ 10\frac{5}{12}.\]

\[2)\ y = x^{2} + 1;\ \ \ y = 3 - x^{2}\]

\[x^{2} + 1 = 3 - x^{2}\]

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

\[x^{2} = 1\ \]

\[x = \pm 1.\]

\[S =\]

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

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

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

\[= 2 - \frac{2}{3} + 2 - \frac{2}{3} =\]

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

\[Ответ:\ \ 2\frac{2}{3}.\]

\[3)\ y = x^{2};\ \ \ y = 2\sqrt{2x}\]

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

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

\[x^{4} - 8x = 0\]

\[x \bullet \left( x^{3} - 8 \right) = 0\]

\[x_{1} = 0\ \ и\ \ x_{2} = 2.\]

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

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

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

\[= \frac{2}{3}\sqrt{4^{3}} - \frac{8}{3} = \frac{2}{3}\sqrt{64} - \frac{8}{3} =\]

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

\[Ответ:\ \ 2\frac{2}{3}.\]

\[4)\ y = \sqrt{x},\ \ \ y = \sqrt{4 - 3x},\ \ \ y = 0\]

\[\sqrt{x} = \sqrt{4 - 3x}\]

\[x = 4 - 3x\]

\[4x = 4\]

\[x = 1.\]

\[\textbf{а)}\ \sqrt{x} > 0\ \]

\[x \neq 0.\]

\[\textbf{б)}\ \sqrt{4 - 3x} > 0\]

\[4 - 3x \neq 0\]

\[3x \neq 4\ \]

\[x \neq \frac{4}{3}.\]

\[S = \int_{0}^{1}{x^{\frac{1}{2}}\text{\ dx}} + \int_{1}^{\frac{4}{3}}{(4 - 3x)^{\frac{1}{2}}\text{\ dx}} =\]

\[= \frac{2}{3} - 0 - \frac{2}{9}\sqrt{0^{3}} + \frac{2}{9}\sqrt{1^{3}} =\]

\[= \frac{2}{3} + \frac{2}{9} = \frac{6 + 2}{9} = \frac{8}{9}.\]

\[Ответ:\ \ \frac{8}{9}.\]