Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1001

\[\boxed{\mathbf{1001}\mathbf{.}}\]

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

\[F(x) = 4 \bullet \frac{x^{1}}{1} - \frac{x^{3}}{3} =\]

\[= 4x - \frac{x^{3}}{3} + C.\]

\[Пересечения\ с\ осью\ x:\]

\[4 - x^{2} > 0\]

\[x^{2} < 4\]

\[- 2 < x < 2.\]

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

\[= F(2) - F( - 2);\]

\[Ответ:\ \ 10\frac{2}{3}.\]

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

\[F(x) = 1 \bullet \frac{x^{1}}{1} - \frac{x^{3}}{3} = x - \frac{x^{3}}{3} + C.\]

\[Пересечения\ с\ осью\ x:\]

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

\[x^{2} < 1\]

\[- 1 < x < 1.\]

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

\[= F(1) - F( - 1);\]

\[S = 1 - \frac{1^{3}}{3} - ( - 1) + \frac{( - 1)^{3}}{3} =\]

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

\[Ответ:\ \ 1\frac{1}{3}.\]

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

\[F(x) = - \frac{x^{3}}{3} + 4 \bullet \frac{x^{2}}{2} - 3 \bullet \frac{x^{1}}{1} =\]

\[= - \frac{x^{3}}{3} + 2x^{2} - 3x + C.\]

\[Пересечения\ с\ осью\ x:\]

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

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

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\text{\ \ }x_{2} = \frac{4 + 2}{2} = 3.\]

\[(x - 1)(x - 3) < 0\]

\[1 < x < 3.\]

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

\[= F(3) - F(1);\]

\[S = - 9 + 18 - 9 + \frac{1}{3} - 2 + 3 =\]

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

\[Ответ:\ \ 1\frac{1}{3}.\]