ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 398

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

\[x = 0;\ \ \ y = 6;\text{\ \ \ x} = 1:\]

\[|S| =\]

\[= \int_{0}^{1}{\left( x^{3} - 3x^{2} - 9x + 1 - 6 \right)\text{dx}} =\]

\[= \int_{0}^{1}{\left( x^{3} - 3x^{2} - 9x - 5 \right)\text{\ dx}} =\]

\[= \left. \ \left( \frac{x^{4}}{4} - 3 \bullet \frac{x^{3}}{3} - 9 \bullet \frac{x^{2}}{2} - 5x \right) \right|_{0}^{1} =\]

\[= \left. \ \left( \frac{x^{4}}{4} - x^{3} - \frac{9}{2}x^{2} - 5x \right) \right|_{0}^{1} =\]

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

\[= - \frac{17}{4} - 6 = - 4\frac{1}{4} - 6 = - 10\frac{1}{4}.\]

\[Ответ:\ \ 10\frac{1}{4}.\]

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

\[y = 1;\ \ \ x = 0;\text{\ \ \ x} = 1:\]

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

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

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

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

\[= - \frac{7}{15} + 4 = 3\frac{8}{15}.\]

\[Ответ:\ \ 3\frac{8}{15}.\]