Страница 316 номер 1039, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(y = x^{2} - 6x + 9\), \(y = x^{2} + 4x + 4\),
\(y = 0;\)
\(x^{2} - 6x + 9 = x^{2} + 4x + 4,\)
\(10x = 5, x = \dfrac{1}{2};\)
\(x_{1} = \dfrac{6 \pm \sqrt{36 - 36}}{2} = 3,\)
\(x_{2} = \dfrac{-4 \pm \sqrt{16 - 16}}{2} = - 2;\)
\(S = S_{1} + S_{2} = \int_{-2}^{\frac{1}{2}}{\left( x^{2} + 4x + 4 \right)dx} +\)
\(+ \int_{\frac{1}{2}}^{3}{\left( x^{2} - 6x + 9 \right)dx} =\)
\(= \left. \ \left( \dfrac{x^{3}}{3} + 2x^{2} + 4x \right) \right|_{-2}^{\frac{1}{2}} + \left. \ \left( \dfrac{x^{3}}{3} - 3x^{2} + 9x \right) \right|_{\frac{1}{2}}^{3} =\)
\(= \dfrac{1}{24} + \dfrac{1}{2} + 2 + \dfrac{8}{3} - 8 + 8 + 9 - 27 + 27 - \dfrac{1}{24} + \dfrac{3}{4} - \dfrac{9}{2} = 10\dfrac{5}{12};\)

2) \(y = x^{2} + 1\), \(y = 3 - x^{2}\)
\(x^{2} + 1 = 3 - x^{2},\)
\(2x^{2} = 2,\)
\(x = \pm 1;\)
\(S = S_{1} - S_{2} = \int_{-1}^{1}{\left( 3 - x^{2} - x^{2} - 1 \right)dx} =\)
\(= \left. \ \left( 2x - \dfrac{2x^{3}}{3} \right) \right|_{-1}^{1} = 2 - \dfrac{2}{3} + 2 - \dfrac{2}{3} =\)
\(= \dfrac{8}{3} = 2\dfrac{2}{3};\)

3) \(y = x^{2}\), \(y = 2\sqrt{2x}\)
\(x^{2} = 2\sqrt{2x},\)
\(x^{4} = 8x,\)
\(x\left( x^{3} - 8 \right) = 0,\)
\(x_{1} = 0, x_{2} = 2;\)
\(S = S_{1} - S_{2} = \int_{0}^{2}{\left( 2\sqrt{2x} - x^{2} \right)dx} =\)
\(= \left. \ \left( \dfrac{2}{3}(2x)^{\frac{3}{2}} - \dfrac{x^{3}}{3} \right) \right|_{0}^{2} = \dfrac{16}{3} - \dfrac{8}{3} = \dfrac{8}{3} = 2\dfrac{2}{3};\)

4) \(y = \sqrt{x}\), \(y = \sqrt{4 - 3x}\), \(y = 0\)
\(\sqrt{x} = \sqrt{4 - 3x}, x = 4 - 3x, x = 1;\)
\(\sqrt{4 - 3x} = 0, 4 - 3x = 0, x = \dfrac{4}{3};\)
\(S = S_{1} + S_{2} = \int_{0}^{1}{\sqrt{x}dx} + \int_{1}^{\frac{4}{3}}{\sqrt{4 - 3x}dx} =\)
\(= \left. \ \dfrac{2}{3}x^{\frac{3}{2}} \right|_{0}^{1} - {\dfrac{1}{3} \cdot \dfrac{2}{3}\left. \ (4 - 3x)^{\frac{3}{2}} \right|}_{1}^{\frac{4}{3}} = \dfrac{2}{3} + \dfrac{2}{9} = \dfrac{8}{9}.\)