ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1100

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

\[\cos x = 0\]

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

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

\[Первая\ трапеция:\]

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

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

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

\[Вторая\ трапеция:\]

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

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

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

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

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

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

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

\[Ответ:\ \ \frac{8}{3\ln 3}.\]

\[3)\ y = 2\cos{3x} - 5\sin{2x} + 10;\]

\[y = 0;\ \ \ x = - \frac{3\pi}{4};\ \ \ x = \frac{5\pi}{4}:\]

\[S =\]

\[= \int_{- \frac{3\pi}{4}}^{\frac{5\pi}{4}}{\left( 2\cos{3x} - 5\sin{2x} + 10 \right)\text{dx}} =\]

\[= \left. \ \left( \frac{2}{3}\sin{3x} + \frac{5}{2}\cos{2x} + 10x \right) \right|_{- \frac{3\pi}{4}}^{\frac{5\pi}{4}} =\]

\[= \frac{40\pi}{2} = 20\pi.\]

\[Ответ:\ \ 20\pi.\]