Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1539

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\[1)\ y = 4x - x^{2};\ y = 5;\]

\[x = 0;\text{\ \ \ x} = 3:\]

\[S = \int_{0}^{3}\left( 4x - x^{2} - 5 \right) =\]

\[= \left. \ \left( 4 \bullet \frac{x^{2}}{2} - \frac{x^{3}}{3} - 5 \bullet \frac{x^{1}}{1} \right) \right|_{0}^{3} =\]

\[= \left. \ \left( 2x^{2} - \frac{x^{3}}{3} - 5x \right) \right|_{0}^{3} =\]

\[= 2 \bullet 3^{2} - \frac{3^{3}}{3} - 5 \bullet 3 - 2 \bullet 0^{2} + \frac{0^{3}}{3} + 5 \bullet 0 =\]

\[= 2 \bullet 9 - \frac{27}{3} - 15 =\]

\[= 18 - 9 - 15 = - 6.\]

\[Ответ:\ \ 6.\]

\[2)\ y = x^{2} - 2x + 8;\ y = 6;\]

\[x = - 1;\text{\ \ \ x} = 3:\]

\[S = \int_{- 1}^{3}\left( x^{2} - 2x + 8 - 6 \right) =\]

\[= \int_{- 1}^{3}\left( x^{2} - 2x + 2 \right) =\]

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

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

\[= \frac{3^{3}}{3} - 3^{2} + 2 \bullet 3 - \frac{( - 1)^{3}}{3} + ( - 1)^{2} - 2 \bullet ( - 1) =\]

\[= \frac{27}{3} - 9 + 6 + \frac{1}{3} + 1 + 2 =\]

\[= \frac{27}{3} + \frac{1}{3} = \frac{28}{3} = 9\frac{1}{3}.\]

\[Ответ:\ \ 9\frac{1}{3}.\]

\[3)\ y = \sin x;\text{\ \ \ y} = 0;\text{\ \ \ }\]

\[x = \frac{2\pi}{3};\ \ \ x = \pi:\]

\[S = \int_{\frac{2\pi}{3}}^{\pi}\left( \sin x \right) = \left. \ \left( - \cos x \right) \right|_{\frac{2\pi}{3}}^{\pi} =\]

\[= - \cos\pi + \cos\frac{2\pi}{3} =\]

\[= - ( - 1) + \cos\left( \pi - \frac{\pi}{3} \right) =\]

\[= 1 - \cos\frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]

\[4)\ y = \cos x;\ \ \ y = 0;\text{\ \ \ }\]

\[x = - \frac{\pi}{6};\ \ \ x = \frac{\pi}{6}:\]

\[S = \int_{- \frac{\pi}{6}}^{\frac{\pi}{6}}\left( \cos x \right) = \left. \ \left( \sin x \right) \right|_{- \frac{\pi}{6}}^{\frac{\pi}{6}} =\]

\[= \sin\frac{\pi}{6} - \sin\left( - \frac{\pi}{6} \right) =\]

\[= \sin\frac{\pi}{6} + \sin\frac{\pi}{6} =\]

\[= \frac{1}{2} + \frac{1}{2} = 1.\]

\[Ответ:\ \ 1.\]