Страница 316 номер 1038, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(y = \frac{1}{x}\), \(y = 4x\), \(x = 1\), \(y = 0\)
\(\dfrac{1}{x} = 4x, 4x^{2} = 1, x = \pm \dfrac{1}{2};\)
\(x = -\frac{1}{2}\) не подходит: по рисунку фигура лежит при \(x > 0\);
\(S = S_{1} + S_{2} =\)
\(= \int_{0}^{\frac{1}{2}}{4xdx} + \int_{\frac{1}{2}}^{1}\dfrac{dx}{x} = \left. \ 2x^{2} \right|_{0}^{\frac{1}{2}} + \left. \ \ln x \right|_{\frac{1}{2}}^{1} =\)
\(= \dfrac{1}{2} + \ln 1 - \ln\dfrac{1}{2} = \dfrac{1}{2} - \ln\dfrac{1}{2};\)

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

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

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