ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 391

\[1)\ y = \sqrt{x};x = 1;x = 4;y = 0:\]

\[|S| = \int_{1}^{4}{\sqrt{x}\text{\ dx}} = \left. \ \frac{2}{3}x^{\frac{3}{2}} \right|_{1}^{4} =\]

\[= \frac{2}{3} \bullet 4^{\frac{3}{2}} - \frac{2}{3} \bullet 1^{\frac{3}{2}} = \frac{2}{3} \bullet 2^{3} - \frac{2}{3} =\]

\[= \frac{16}{3} - \frac{2}{3} = \frac{14}{3} = 4\frac{2}{3}.\]

\[Ответ:\ \ 4\frac{2}{3}.\]

\[2)\ y = \cos x;x = 0;x = \frac{\pi}{3};y = 0:\]

\[|S| = \int_{0}^{\frac{\pi}{3}}{\cos x\text{dx}} = \left. \ - \sin x \right|_{0}^{\frac{\pi}{3}} =\]

\[= - \sin\frac{\pi}{3} + \sin 0 = - \frac{\sqrt{3}}{2}.\]

\[Ответ:\ \ \frac{\sqrt{3}}{2}.\]

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

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

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

\[D = 1 + 8 = 9\]

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

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

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

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

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

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

\[Ответ:\ \ 4\frac{1}{2}.\]

\[4)\ y = 2x^{2};\ \ \ y = 0,5x + 1,5:\]

\[2x^{2} = 0,5x + 1,5\]

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

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 - 7}{2 \bullet 4} = - \frac{3}{4};\text{\ \ }\]

\[x_{2} = \frac{1 + 7}{2 \bullet 4} = 1.\]

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

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

\[= \frac{2}{3} - \frac{1}{4} - \frac{3}{2} + \frac{18}{64} + \frac{9}{64} - \frac{9}{8} =\]

\[= \frac{128 - 48 - 288 + 54 + 27 - 216}{192} =\]

\[= - \frac{343}{192} = - 1\frac{151}{192}.\]

\[Ответ:\ \ 1\frac{151}{192}.\]

\[5)\ y = \sqrt[3]{x};\ x = - 8;x = - 1;y = 0:\]

\[|S| = \int_{- 8}^{- 1}{\sqrt[3]{x}\text{\ dx}} = \left. \ \frac{3}{4}x^{\frac{4}{3}} \right|_{- 8}^{- 1} =\]

\[= \frac{3}{4} \bullet ( - 1)^{\frac{4}{3}} - \frac{3}{4} \bullet ( - 8)^{\frac{4}{3}} =\]

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

\[= - 15 \bullet \frac{3}{4} = - \frac{45}{4} = - 11\frac{1}{4}.\]

\[Ответ:\ \ 11\frac{1}{4}.\]

\[6)\ y = \frac{1}{x^{3}};x = - 3;x = - 1;y = 0:\]

\[|S| = \int_{- 3}^{- 1}{\frac{1}{x^{3}}\text{dx}} = \left. \ \frac{x^{- 2}}{- 2} \right|_{- 3}^{- 1} =\]

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

\[= \frac{- 9 + 1}{18} = - \frac{8}{18} = - \frac{4}{9}.\]

\[Ответ:\ \ \frac{4}{9}.\]