Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 75

\[\boxed{\mathbf{75}.}\]

\[\textbf{а)}\ Рисунок\ в\ учебнике.\]

\[x^{2} + 9y^{2} = 9\]

\[Верхняя\ граница\ эллипса:\]

\[y = \sqrt{1 - \left( \frac{x}{3} \right)^{2}};\]

\[S_{1} = \int_{- 3}^{3}{\sqrt{1 - \left( \frac{x}{3} \right)^{2}}\text{dx}};\]

\[x \in \lbrack - 3;3\rbrack:\]

\[\frac{x}{3} = \sin u;\ \ u \in \left\lbrack - \frac{\pi}{2};\frac{\pi}{2} \right\rbrack;\]

\[\cos u \geq 0;\ \ dx = 3\cos udu:\]

\[S_{1} = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{3\cos^{2}u\text{du}} =\]

\[= 3\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{1 + \cos{2u}}{2}\text{du}} =\]

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

\[2S_{1} = 3\pi.\]

\[Ответ:3\pi.\]

\[\textbf{б)}\ Рисунок\ в\ учебнике.\]

\[4x^{2} + y^{2} = 4\]

\[Верхняя\ граница\ эллипса:\]

\[y = \sqrt{2^{2} - {(2x)}^{2}} = 2\sqrt{1 - x^{2}}.\]

\[Найдем\ плошадь\ \frac{1}{4}\ эллипса:\]

\[S_{\frac{1}{4}} = 2\int_{0}^{1}{\sqrt{1 - x^{2}}\text{dx}} =\]

\[= 2\int_{0}^{3}{\sqrt{1 - x^{2}}\text{dx}};\]

\[x = \sin t;dx = \cos tdt;\]

\[2\int_{0}^{3}{\sqrt{1 - x^{2}}\text{dx}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\sqrt{1 - {(\sin t)}^{2}}\cos t\text{dt}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\sqrt{1 - sin^{2}t}\cos t\text{dt}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\sqrt{\text{co}s^{2}t}\cos t\text{dt}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\left| \cos t \right|\cos t\text{dt}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\cos^{2}t\text{dt}} =\]

\[= 2\int_{0}^{\frac{\pi}{2}}{\frac{1 + \cos{2t}}{2}\text{dt}} =\]

\[= \frac{2}{2}2\int_{0}^{\frac{\pi}{2}}{\left( 1 + \cos{2t} \right)\text{dt}} =\]

\[= \left. \ t + \frac{\sin{2t}}{2} \right|_{0}^{\frac{\pi}{2}} =\]

\[S = 4 \cdot S_{\frac{1}{4}} = 4 \cdot \frac{\pi}{2} = 2\pi.\]

\[Ответ:2\pi.\]