Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 376

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

\[1 - x^{2} = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

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

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

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

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

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

\[Ответ:\ \ 1\frac{1}{3}.\]

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

\[- x^{2} + 4x - 3 = 0\]

\[x^{2} - 4x + 3 = 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\]

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

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

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

\[= ( - 9 + 18 - 9) - \left( - \frac{1}{3} + 2 - 3 \right) =\]

\[= 1 + \frac{1}{3} = 1\frac{1}{3}.\]

\[Ответ:\ \ 1\frac{1}{3}.\]

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

\[- x(x + 3) = 0\]

\[x_{1} = - 3;\text{\ \ \ }x_{2} = 0.\]

\[S = \int_{- 3}^{0}{- x(x + 3)\text{dx}} =\]

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

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

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

\[= - \left( 9 - \frac{27}{2} \right) = 13\frac{1}{2} - 9 = 4\frac{1}{2}.\]

\[Ответ:\ \ 4\frac{1}{2}.\]

\[4)\ y = (1 - x)(x + 2)\]

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

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

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

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

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

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

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

\[Ответ:\ \ 4\frac{1}{2}.\]

\[5)\ y = (x + 2)(3 - x)\]

\[(x + 2)(3 - x) = 0\]

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

\[S = \int_{- 2}^{3}{(x + 2)(3 - x)\text{dx}} =\]

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

\[= \left. \ \left( - \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6x \right) \right|_{- 2}^{3} =\]

\[= \left( - 9 + \frac{9}{2} + 18 \right) - \left( \frac{8}{3} + \frac{4}{2} - 12 \right) =\]

\[= 21 + \frac{5}{2} - \frac{8}{3} = 21 + \frac{15 - 16}{6} =\]

\[= 20\frac{5}{6}.\]

\[Ответ:\ \ 20\frac{5}{6}.\]