Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 78

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\[y = \sin x;\ \ 0 \leq x \leq \pi;\]

\[V = \pi\int_{a}^{b}{\left( f(x) \right)^{2}\text{dx}} =\]

\[= \pi\int_{0}^{\pi}{\left( \sin x \right)^{2}\text{dx}} =\]

\[= \pi\int_{0}^{\pi}{\frac{1 - \cos{2x}}{2}\text{dx}} =\]

\[= \pi \cdot \frac{1}{2}\int_{0}^{\pi}{\left( 1 - \cos{2x} \right)\text{dx}} =\]

\[= \frac{\pi}{2}\int_{0}^{\pi}{\left( 1 - \cos{2x} \right)\text{dx}} =\]

\[= \frac{\pi}{2}\left( \int_{0}^{\pi}{1dx} - \int_{0}^{\pi}{\cos{2x}\text{dx}} \right) =\]

\[= \frac{\pi}{2} \cdot \left( \left. \ x \right|_{0}^{\pi} - \left. \ \frac{1}{2}\sin{2x} \right|_{0}^{\pi} \right) =\]

\[= \frac{\pi}{2}\left( \pi - \frac{1}{2}\left( \sin{2\pi} - \sin 0 \right) \right) =\]

\[= \frac{\pi}{2}\left( \pi - \frac{1}{2}(0 - 0) \right) = \frac{\pi}{2} \cdot \pi =\]

\[= \frac{\pi^{2}}{2}.\]

\[Ответ:V = \frac{\pi^{2}}{2}.\]