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МатематикаРешебник по алгебре и начала математического анализа 11 класс Алимов Задание 1017
Задание 1017
\[\boxed{\mathbf{1017}\mathbf{.}}\]
\[1)\ y = x^{2} + 1\ \ и\ \ y = 3 - x\]
\[x^{2} + 1 = 3 - x\]
\[x^{2} + x - 2 = 0\]
\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]
\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\ \ \]
\[x_{2} = \frac{- 1 + 3}{2} = 1.\]
\[S =\]
\[= \int_{- 2}^{1}{\left( (3 - x) - \left( x^{2} + 1 \right) \right)\text{\ dx}} =\]
\[= \int_{- 2}^{1}{\left( 2 - x^{2} - x \right)\text{\ dx}} =\]
\[= \left. \ \left( 2x - \frac{x^{3}}{3} - \frac{x^{2}}{2} \right) \right|_{- 2}^{1} =\]
\[= 2 - \frac{1}{3} - \frac{1}{2} + 4 - \frac{8}{3} + \frac{4}{2} =\]
\[= 6 - \frac{9}{3} + \frac{3}{2} = 6 - 3 + 1,5 = 4,5.\]
\[Ответ:\ \ 4,5.\]
\[2)\ y = (x + 2)^{2}\text{\ \ }и\ \ y = x + 2\]
\[(x + 2)^{2} = x + 2\]
\[x^{2} + 4x + 4 - x - 2 = 0\]
\[x^{2} + 3x + 2 = 0\]
\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]
\[x_{1} = \frac{- 3 - 1}{2} = - 2\ \ и\ \ \]
\[x_{2} = \frac{- 3 + 1}{2} = - 1.\]
\[S =\]
\[= \int_{- 2}^{- 1}{\left( (x + 2) - (x + 2)^{2} \right)\text{\ dx}} =\]
\[= \left. \ \left( \frac{x^{2}}{2} + 2x - \frac{(x + 2)^{3}}{3} \right) \right|_{- 2}^{- 1} =\]
\[= \frac{1}{2} - 2 - \frac{1^{3}}{3} - \frac{4}{2} + 4 + \frac{0^{3}}{3} =\]
\[= \frac{1}{2} - 2 - \frac{1}{3} - 2 + 4 = \frac{1}{2} - \frac{1}{3} =\]
\[= \frac{3}{6} - \frac{2}{6} = \frac{1}{6}.\]
\[Ответ:\ \ \frac{1}{6}.\]
\[3)\ y = \sqrt{x}\text{\ \ }и\ \ y = x\]
\[\sqrt{x} = x\]
\[x = x^{2}\]
\[x^{2} - x = 0\]
\[x \bullet (x - 1) = 0\]
\[x_{1} = 0\ и\ x_{2} = 1.\]
\[S = \int_{0}^{1}{\left( \sqrt{x} - x \right)\text{\ dx}} =\]
\[= \int_{0}^{1}{\left( x^{\frac{1}{2}} - x \right)\text{\ dx}} =\]
\[= \left. \ \left( x^{\frac{3}{2}}\ :\frac{3}{2} - \frac{x^{2}}{2} \right) \right|_{0}^{1} =\]
\[= \left. \ \left( \frac{2}{3}x\sqrt{x} - \frac{x^{2}}{2} \right) \right|_{0}^{1} =\]
\[= \frac{2}{3} \bullet 1\sqrt{1} - \frac{1^{2}}{2} - \frac{2}{3} \bullet 0\sqrt{0} + \frac{0^{2}}{2} =\]
\[= \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}.\]
\[Ответ:\ \ \frac{1}{6}.\]
