ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1116

\[f(x) = \sin x;\ \ \ y = 0;\text{\ \ \ }\]

\[x = \frac{\pi}{2}\ \left( 0 \leq x \leq \frac{\pi}{2} \right):\]

\[1)\ Пересечение\ с\ осью\ Ox:\]

\[\sin x = 0;\ \ \ x = \pi n;\text{\ \ \ x} = 0.\]

\[2)\ S = \int_{0}^{\frac{\pi}{2}}{\sin x\text{\ dx}} = \left. \ - \cos x \right|_{0}^{\frac{\pi}{2}} =\]

\[= - \cos\frac{\pi}{2} + \cos 0 = - 0 + 1 = 1.\]

\[3)\ Уравнение\ прямой:\]

\[y = kx = tg\ a \bullet x.\]

\[4)\ Площадь\ половины\ фигуры:\]

\[S_{1} = \int_{0}^{\frac{\pi}{2}}(tg\ a \bullet x) =\]

\[= \left. \ \left( tg\ a \bullet \frac{x^{2}}{2} \right) \right|_{0}^{\frac{\pi}{2}} =\]

\[= tg\ a \bullet \left( \frac{\pi}{2} \right)^{2}:\ 2 =\]

\[= tg\ a \bullet \frac{\pi^{2}}{8} = \frac{1}{2};\]

\[tg\ a = \frac{8}{2\pi^{2}} = \ \frac{4}{\pi^{2}}\]

\[a = arctg\frac{4}{\pi^{2}}.\]

\[Ответ:\ \ arctg\frac{4}{\pi^{2}}.\]