Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1541

\[\boxed{\mathbf{1541}\mathbf{.}}\]

\[1)\ y = 9 - x^{2};\ y = (x - 1)^{2} - 4:\]

\[9 - x^{2} = (x - 1)^{2} - 4\]

\[9 - x^{2} = x^{2} - 2x + 1 - 4\]

\[9 - x^{2} = x^{2} - 2x - 3\]

\[2x^{2} - 2x - 12 = 0\]

\[x^{2} - x - 6 = 0\]

\[D = 1 + 24 = 25\]

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

\[x_{2} = \frac{1 + 5}{2} = 3.\]

\[S = \int_{- 2}^{3}\left( 9 - x^{2} - x^{2} + 2x + 3 \right) =\]

\[= \int_{- 2}^{3}\left( 12 - 2x^{2} + 2x \right) =\]

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

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

\[= 36 - 2 \bullet 9 + 9 + 24 - \frac{2 \bullet 8}{3} - 4 =\]

\[= 65 - 18 - \frac{16}{3} =\]

\[= 47 - 5\frac{1}{3} = 41\frac{2}{3}.\]

\[Ответ:\ \ \ 41\frac{2}{3}.\]

\[2)\ y = x^{2};\ \text{\ \ }y = \sqrt[3]{x}:\]

\[x^{2} = \sqrt[3]{x}\]

\[\left( x^{2} \right)^{3} = x\]

\[x^{6} = x\]

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

\[S = \int_{0}^{1}\left( x^{2} - \sqrt[3]{x} \right) =\]

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

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

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

\[= \frac{1^{3}}{3} - \frac{3\sqrt[3]{1^{4}}}{4} - \frac{0^{3}}{3} + \frac{3\sqrt[3]{0^{4}}}{4} =\]

\[= \frac{1}{3} - \frac{3}{4} = \frac{4 - 3 \bullet 3}{12} =\]

\[= \frac{4 - 9}{12} = - \frac{5}{12}.\]

\[Ответ:\ \ \frac{5}{12}.\]