Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 383

\[1)\ y = - x^{2} + 4x - 3;\]

\[прямая\ (1;\ 0)\ и\ (0;\ - 3):\]

\[0 = k \bullet 1 + b\ \ \ \]

\[b = - 3\]

\[- 3 = k \bullet 0 + b\text{\ \ }\]

\[k = - b = 3.\]

\[Точки\ пересечения:\]

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

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

\[x(x - 1) = 0\]

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

\[|S| =\]

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

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

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

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

\[= \frac{3 - 2}{6} = \frac{1}{6}.\]

\[Ответ:\ \ \frac{1}{6}.\]

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

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

\[x^{2} = 2\]

\[x = \pm \sqrt{2}.\]

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

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

\[= \left( - \frac{2\sqrt{2}}{3} + 2\sqrt{2} \right) - \left( \frac{2\sqrt{2}}{3} - 2\sqrt{2} \right) =\]

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

\[= \frac{- 4\sqrt{2} + 12\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}.\]

\[Ответ:\ \ \frac{8\sqrt{2}}{3}.\]

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

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

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

\[x^{2} = 1\]

\[x = \pm 1.\]

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

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

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

\[= \left( 2 - \frac{2}{3} \right) - \left( - 2 + \frac{2}{3} \right) =\]

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

\[Ответ:\ \ 2\frac{2}{3}.\]

\[4)\ y = x^{3};\ y = 1;\ x = - 2:\]

\[x^{3} = 1\]

\[x = 1.\]

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

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

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

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

\[Ответ:\ \ 6\frac{3}{4}.\]

\[5)\ y = x;\ y = x^{3}\ на\ \lbrack - 1;\ 0\rbrack:\]

\[|S| = \int_{- 1}^{0}{\left( x^{3} - x \right)\text{dx}} =\]

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

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

\[Ответ:\ \ \frac{1}{4}.\]

\[6)\ y = x^{2} - 2x;\ y = - x^{2}:\]

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

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

\[2x(x - 1) = 0\]

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

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

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

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

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

\[Ответ:\ \ \frac{1}{3}.\]