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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 1035
Задание 1035
\[\boxed{\mathbf{1035}\mathbf{.}}\]
\[1)\ y = \sqrt{x},\ \ \ x = 1,\ \ \ x = 4,\ \ \ y = 0.\]
\[Пересечения\ с\ осью\ x:\]
\[\sqrt{x} > 0\]
\[x \neq 0.\]
\[S = \int_{1}^{4}{\sqrt{x}\text{\ dx}} = \int_{1}^{4}{x^{\frac{1}{2}}\text{\ dx}} =\]
\[= \left. \ \left( x^{\frac{3}{2}}\ :\frac{3}{2} \right) \right|_{1}^{4} = \left. \ \left( \frac{2}{3}x\sqrt{x} \right) \right|_{1}^{4};\]
\[S = \frac{2}{3} \bullet 4\sqrt{4} - \frac{2}{3} \bullet 1\sqrt{1} =\]
\[= \frac{8}{3} \bullet 2 - \frac{2}{3} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3} = 4\frac{2}{3}.\]
\[Ответ:\ \ 4\frac{2}{3}.\]
\[2)\ y = \cos x,\ \ \ x = 0,\ \ \ x = \frac{\pi}{3},\ \ \ \]
\[y = 0.\]
\[Пересечения\ с\ осью\ x:\]
\[\cos x > 0\]
\[- \frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n\]
\[0 < x < \frac{\pi}{3}.\]
\[S = \int_{0}^{\frac{\pi}{3}}{\cos x\text{\ dx}} = \left. \ \sin x\ \right|_{0}^{\frac{\pi}{3}} =\]
\[= \sin\frac{\pi}{3} - \sin 0 = \frac{\sqrt{3}}{2}.\]
\[Ответ:\ \ \frac{\sqrt{3}}{2}.\]
\[3)\ y = x^{2}\text{\ \ }и\ \ y = 2 - x\]
\[x^{2} = 2 - x\]
\[x^{2} + x - 2 = 0\]
\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]
\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\ \ \]
\[x_{2} = \frac{- 1 + 3}{2} = 1.\]
\[S = \int_{- 2}^{1}{\left( 2 - x - x^{2} \right)\text{\ dx}} =\]
\[= \left. \ \left( 2x - \frac{x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{- 2}^{1} =\]
\[= 2 - \frac{1}{2} - \frac{1}{3} + 4 + \frac{4}{2} - \frac{8}{3} =\]
\[= 6 + \frac{3}{2} - \frac{9}{3} = 6 + 1,5 - 3 = 4,5.\]
\[Ответ:\ \ 4,5.\]
\[4)\ y = 2x^{2}\text{\ \ }и\ \ y = 0,5x + 1,5\]
\[2x^{2} = 0,5x + 1,5\]
\[2x^{2} - 0,5x - 1,5 = 0\]
\[4x^{2} - x - 3 = 0\]
\[D = 1^{2} + 4 \bullet 4 \bullet 3 = 1 + 48 = 49\]
\[x_{1} = \frac{1 - 7}{2 \bullet 4} = - \frac{6}{8} = - \frac{3}{4}\text{\ \ }и\ \ \]
\[x_{2} = \frac{1 + 7}{2 \bullet 4} = 1.\]
\[S = \int_{- \frac{3}{4}}^{1}{\left( 0,5x + 1,5 - 2x^{2} \right)\text{\ dx}} =\]
\[= \left. \ \left( 0,5 \bullet \frac{x^{2}}{2} + 1,5x - 2 \bullet \frac{x^{3}}{3} \right) \right|_{- \frac{3}{4}}^{1} =\]
\[= \frac{1}{4} + \frac{3}{2} - \frac{2}{3} - \frac{9}{64} + \frac{9}{8} - \frac{27}{96} =\]
\[= \frac{343}{192} = 1\frac{151}{192}.\]
\[Ответ:\ \ 1\frac{151}{192}.\]
