ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1038

\[\boxed{\mathbf{1038}\mathbf{.}}\]

\[1)\ y = \frac{1}{x},\ \ \ y = 4x,\ \ \ x = 1,\ \ \ \]

\[y = 0.\]

\[\frac{1}{x} = 4x\]

\[1 = 4x^{2}\]

\[x^{2} = \frac{1}{4}\ \]

\[x = \pm \frac{1}{2}.\]

\[\textbf{а)}\ \frac{1}{x} > 0\ \]

\[x > 0.\]

\[\textbf{б)}\ 4x > 0\ \]

\[x > 0.\]

\[S = \int_{0}^{0,5}{4x\ dx} + \int_{0,5}^{1}{\frac{1}{x}\text{\ dx}} =\]

\[= \left. \ \left( 4 \bullet \frac{x^{2}}{2} \right) \right|_{0}^{0,5} + \left. \ \ln x \right|_{0,5}^{1} =\]

\[= \left. \ 2x^{2} \right|_{0}^{0,5} + \left. \ \ln x \right|_{0,5}^{1} =\]

\[= 2 \bullet {0,5}^{2} - 2 \bullet 0^{2} + \ln 1 - \ln{0,5} =\]

\[= 2 \bullet 0,25 - 0 + 0 - \ln{0,5} =\]

\[= \frac{1}{2} - \ln\frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2} - \ln\frac{1}{2}.\]

\[2)\ y = \frac{1}{x^{2}},\ \ \ y = x,\ \ \ x = 2,\ \ \ \]

\[y = 0:\]

\[\frac{1}{x^{2}} = x\]

\[1 = x^{3}\]

\[x = 1.\]

\[\textbf{а)}\ \frac{1}{x^{2}} > 0\ \]

\[x \neq 0.\]

\[\textbf{б)}\ x > 0.\]

\[S = \int_{0}^{1}\text{x\ dx} + \int_{1}^{2}{x^{- 2}\text{\ dx}} =\]

\[= \left. \ \frac{x^{2}}{2} \right|_{0}^{1} + \left. \ \left( x^{- 1}\ :( - 1) \right) \right|_{1}^{2} =\]

\[= \left. \ \frac{x^{2}}{2} \right|_{0}^{1} + \left. \ \left( - \frac{1}{x} \right) \right|_{1}^{2} =\]

\[= \frac{1^{2}}{2} - \frac{0^{2}}{2} + \left( - \frac{1}{2} \right) - \left( - \frac{1}{1} \right) =\]

\[= \frac{1}{2} - 0 - \frac{1}{2} + 1 = 1.\]

\[Ответ:\ \ 1.\]

\[3)\ y = x^{2} + 1\ \ и\ \ y = x + 1\]

\[x^{2} + 1 = x + 1\]

\[x^{2} - x = 0\]

\[x \bullet (x - 1) = 0\]

\[x_{1} = 0\ и\ x_{2} = 1.\]

\[S = \int_{0}^{1}{\left( (x + 1) - \left( x^{2} + 1 \right) \right)\text{\ dx}} =\]

\[= \left. \ \left( \frac{x^{2}}{2} + x - \left( \frac{x^{3}}{3} + x \right) \right) \right|_{0}^{1} =\]

\[= \left. \ \left( \frac{x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{0}^{1};\]

\[S = \frac{1^{2}}{2} - \frac{1^{3}}{3} - \frac{0^{2}}{2} + \frac{0^{3}}{3} = \frac{1}{2} - \frac{1}{3} =\]

\[= \frac{3 - 2}{6} = \frac{1}{6}.\]

\[Ответ:\ \ \frac{1}{6}.\]

\[4)\ y = x^{2} + 2\ \ и\ \ y = 2x + 2\]

\[x^{2} + 2 = 2x + 2\]

\[x^{2} - 2x = 0\]

\[x \bullet (x - 2) = 0\]

\[x_{1} = 0\ и\ x_{2} = 2.\]

\[S =\]

\[= \int_{0}^{2}{\left( (2x + 2) - \left( x^{2} + 2 \right) \right)\text{\ dx}} =\]

\[= \int_{0}^{2}{\left( 2x - x^{2} \right)\text{\ dx}} =\]

\[{= \left. \ \left( 2 \bullet \frac{x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{0}^{2} = }{= \left. \ \left( x^{2} - \frac{x^{3}}{3} \right) \right|_{0}^{2} =}\]

\[= 2^{2} - \frac{2^{3}}{3} - 0^{2} + \frac{0^{3}}{3} = 4 - \frac{8}{3} =\]

\[= 4 - 2\frac{2}{3} = 1\frac{1}{3}.\]

\[Ответ:\ \ 1\frac{1}{3}.\]