ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 377

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

\[S =\]

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

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

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

\[= \frac{3 + 2}{6} = \frac{5}{6}.\]

\[Ответ:\ \ \frac{5}{6}.\]

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

\[S =\]

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

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

\[= 10 - \frac{7}{3} - \frac{3}{2} = 10 - \frac{14 + 9}{6} =\]

\[= 10 - \frac{23}{6} = 10 - 3\frac{5}{6} = 6\frac{1}{6}.\]

\[Ответ:\ \ 6\frac{1}{6}.\]

\[3)\ y = 4x - x^{2};\ (4;\ 0)\ и\ (1;\ 3):\]

\[Уравнение\ прямой:\]

\[0 = k \bullet 4 + b\ \ \ \]

\[b = - 4k.\]

\[3 = k \bullet 1 + b\text{\ \ \ }\]

\[3 = k - 4k\]

\[3k = - 3\]

\[k = - 1;\text{\ \ }\]

\[b = 4.\]

\[S =\]

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

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

\[= 14 - \frac{1}{3} - \frac{15}{2} = = 14 - \frac{2 + 45}{6} =\]

\[= 14 - \frac{47}{6} = 14 - 7\frac{5}{6} = 6\frac{1}{6}.\]

\[Ответ:\ \ 6\frac{1}{6}.\]

\[4)\ y = 3x^{2};\ \ ( - 3;\ 0)\ и\ ( - 1;\ 3):\]

\[Уравнение\ прямой:\]

\[0 = k \bullet ( - 3) + b,\ \ \ \]

\[b = 3k.\]

\[3 = k \bullet ( - 1) + b\text{\ \ }\]

\[3 = - k + 3k\]

\[2k = 3\]

\[k = \frac{3}{2};\text{\ \ \ }\]

\[b = \frac{9}{2}.\]

\[S =\]

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

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

\[= 1 + \left( \frac{3}{4} - \frac{9}{2} \right) - \left( \frac{27}{4} - \frac{27}{2} \right) =\]

\[= 1 - \frac{24}{4} + \frac{18}{2} = 1 - 6 + 9 = 4.\]

\[Ответ:\ \ 4.\]

\[5)\ y = 6x^{2};\ y = (x - 3)(x - 4):\]

\[y = (x - 3)(x - 4) =\]

\[= x^{2} - 4x - 3x + 12 =\]

\[= x^{2} - 7x + 12.\]

\[S =\]

\[= \int_{1}^{3}{\left( x^{2} - 7x + 12 \right)\text{dx}} + \int_{0}^{1}{6x^{2}\text{\ dx}} =\]

\[= \left. \ \left( \frac{1}{3}x^{3} - \frac{7}{2}x^{2} + 12x \right) \right|_{1}^{3} + \left. \ 2x^{3} \right|_{0}^{1} =\]

\[= 35 - \frac{56}{2} - \frac{1}{3} = 7 - \frac{1}{3} = 6\frac{2}{3}.\]

\[Ответ:\ \ 6\frac{2}{3}.\]

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

\[S =\]

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

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

\[= \frac{0^{3}}{3} - \frac{( - 2)^{3}}{3} - \left( - 8 + \frac{8}{3} \right) =\]

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

\[Ответ:\ \ 8.\]