Страница 308 номер 1014, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(y = (x + 1)^{2}\), \(y = 1 - x\), \(Ox\)
\((x + 1)^{2} = 0, x = - 1;\)
\(1 - x = 0, x = 1;\)
\((x + 1)^{2} = 1 - x, x^{2} + 3x = 0, x_{1} = 0, x_{2} = - 3;\)
\(x = - 3\) не подходит: по рисунку фигура лежит при \(- 1 \leqslant x \leqslant 1\);
\(S = S_{1} + S_{2} =\)
\(= \int_{-1}^{0}{(x + 1)^{2}dx} + \int_{0}^{1}{(1 - x)dx} =\)
\(= \int_{-1}^{0}{\left( x^{2} + 2x + 1 \right)dx} + \int_{0}^{1}{(1 - x)dx} =\)
\(= \left. \ \left( \dfrac{x^{3}}{3} + x^{2} + x \right) \right|_{-1}^{0} + \left. \ \left( x - \dfrac{x^{2}}{2} \right) \right|_{0}^{1} =\)
\(= \dfrac{1}{3} - 1 + 1 + 1 - \dfrac{1}{2} = 1 - \dfrac{1}{6} = \dfrac{5}{6};\)

2) \(y = 4 - x^{2}\), \(y = x + 2\), \(Ox\)
\(x + 2 = 0, x = - 2;\)
\(4 - x^{2} = 0, x_{1{,}2} = \pm 2;\)
\(4 - x^{2} = x + 2, x^{2} + x - 2 = 0, x_{1} = 1, x_{2} = - 2;\)
\(S = S_{1} + S_{2} =\)
\(= \int_{-2}^{1}{(x + 2)dx} + \int_{1}^{2}{\left( 4 - x^{2} \right)dx} =\)
\(= \left. \ \left( \dfrac{x^{2}}{2} + 2x \right) \right|_{-2}^{1} + \left. \ \left( 4x - \dfrac{x^{3}}{3} \right) \right|_{1}^{2} =\)
\(= \dfrac{1}{2} + 2 - 2 + 4 + 8 - \dfrac{8}{3} - 4 + \dfrac{1}{3} =\)
\(= 8 - \dfrac{11}{6} = \dfrac{37}{6} = 6\dfrac{1}{6};\)

3) \(y = 4x - x^{2}\), \(y = 4 - x\), \(Ox\)
\(4x - x^{2} = 0, x_{1} = 0, x_{2} = 4;\)
\(4 - x = 0, x = 4;\)
\(4x - x^{2} = 4 - x, x^{2} - 5x + 4 = 0, x_{1} = 1, x_{2} = 4;\)
\(S = S_{1} + S_{2} =\)
\(= \int_{0}^{1}{\left( 4x - x^{2} \right)dx} + \int_{1}^{4}{(4 - x)dx} =\)
\(= \left. \ \left( 2x^{2} - \dfrac{x^{3}}{3} \right) \right|_{0}^{1} + \left. \ \left( 4x - \dfrac{x^{2}}{2} \right) \right|_{1}^{4} =\)
\(= 2 - \dfrac{1}{3} + 16 - 8 - 4 + \dfrac{1}{2} = 6\dfrac{1}{6};\)

4) \(y = 3x^{2}\), \(y = 1{,}5x + 4{,}5\), \(Ox\)
\(1{,}5x + 4{,}5 = 0, x = - 3;\)
\(3x^{2} = 0, x = 0;\)
\(3x^{2} = 1{,}5x + 4{,}5,\)
\(2x^{2} - x - 3 = 0, x_{1} = - 1, x_{2} = 1{,}5;\)
\(x = 1{,}5\) не подходит: по рисунку фигура лежит при \(- 3 \leqslant x \leqslant 0\);
\(S = S_{1} + S_{2} =\)
\(= \int_{-3}^{-1}{(1{,}5x + 4{,}5)dx} + \int_{-1}^{0}{3x^{2}dx} =\)
\(= \left. \ \left( \dfrac{1{,}5x^{2}}{2} + 4{,}5x \right) \right|_{-3}^{-1} + \left. \ x^{3} \right|_{-1}^{0} =\)
\(= \dfrac{1{,}5}{2} - 4{,}5 - \dfrac{13{,}5}{2} + 13{,}5 + 1 = 4.\)