Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1542

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\[1)\ y = \cos x;\ x = \frac{\pi}{4};\text{\ y} = 0:\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n\]

\[x = \frac{\pi}{2} + \pi n;\]

\[x = \pm \frac{\pi}{2} - ближайшие\ точки.\]

\[S_{1} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left( \cos x \right) = \left. \ \left( \sin x \right) \right|_{\frac{\pi}{4}}^{\frac{\pi}{2}} =\]

\[= \sin\frac{\pi}{2} - \sin\frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} =\]

\[= \frac{2 - \sqrt{2}}{2};\]

\[S_{2} = \int_{- \frac{\pi}{2}}^{\frac{\pi}{4}}\left( \cos x \right) = \left. \ \left( \sin x \right) \right|_{- \frac{\pi}{2}}^{\frac{\pi}{4}} =\]

\[= \sin\frac{\pi}{4} - \sin\left( - \frac{\pi}{2} \right) = \frac{\sqrt{2}}{2} + 1 =\]

\[= \frac{2 + \sqrt{2}}{2}.\]

\[Ответ:\ \ \frac{2 - \sqrt{2}}{2};\ \ \frac{2 + \sqrt{2}}{2}.\]

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

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

\[= \frac{3^{1}}{\ln 3} - \frac{3^{- 1}}{\ln 3} = \frac{3 - \frac{1}{3}}{\ln 3} = \frac{8}{3\ln 3}.\]

\[Ответ:\ \ \frac{8}{3\ln 3}.\]