Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1041

\[\boxed{\mathbf{1041}\mathbf{.}}\]

\[1)\ y = x^{3} - 3x^{2} - 9x + 1,\ \ \ \]

\[x = 0,\ \ \ y = 6\ \ (при\ x < 0);\]

\[x^{3} - 3x^{2} - 9x + 1 = 6\]

\[x^{3} - 3x^{2} - 9x - 5 = 0\]

\[x = - 1.\]

\[= 5 + \frac{1}{4} + 1 - \frac{9}{2} = 6 + \frac{1 - 18}{4} =\]

\[= 6 - \frac{17}{4} = 6 - 4\frac{1}{4} = 1\frac{3}{4}.\]

\[Ответ:\ \ 1\frac{3}{4}.\]

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

\[x = 0,\ \ \ x = 1;\]

\[x^{4} - 2x^{2} + 5 = 1\]

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

\[D = 2^{2} - 4 \bullet 4 = 4 - 16 = - 12\]

\[D < 0 \rightarrow корней\ нет.\]

\[S = \int_{0}^{1}{\left( x^{4} - 2x^{2} + 5 - 1 \right)\text{\ dx}} =\]

\[= \int_{0}^{1}{\left( x^{4} - 2x^{2} + 4 \right)\text{\ dx}} =\]

\[= \left. \ \left( \frac{x^{5}}{5} - 2 \bullet \frac{x^{3}}{3} + 4x \right) \right|_{0}^{1} =\]

\[= \frac{1}{5} - \frac{2}{3} + 4 = \frac{3 - 10 + 60}{15} =\]

\[= \frac{53}{15} = 3\frac{8}{15}.\]

\[Ответ:\ \ 3\frac{8}{15}.\]