Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 381

\[1)\ y = 6x - x^{2};\ y = x + 4:\]

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

\[x^{2} - 5x + 4 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2} = 1;\text{\ \ }\]

\[x_{2} = \frac{5 + 3}{2} = 4.\]

\[|S| = \int_{1}^{4}{\left( 6x - x^{2} - x - 4 \right)\text{dx}} =\]

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

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

\[= \left( \frac{80}{2} - \frac{64}{3} - 16 \right) - \left( \frac{5}{2} - \frac{1}{3} - 4 \right) =\]

\[= \frac{75}{2} - \frac{63}{3} - 12 = 37,5 - 33 = 4,5.\]

\[Ответ:\ \ 4,5.\]

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

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

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

\[D = 1 + 8 = 9\]

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

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

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

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

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

\[= \left( 2 - \frac{1}{3} - \frac{1}{2} \right) - \left( - 4 + \frac{8}{3} - \frac{4}{2} \right) =\]

\[= 6 - \frac{9}{3} + \frac{3}{2} = 6 - 3 + 1,5 = 4,5.\]

\[Ответ:\ \ 4,5.\]