Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1015

\[\boxed{\mathbf{1015}\mathbf{.}}\]

\[1)\ y = \sqrt{x}\text{\ \ }и\ \ y = (x - 2)^{2}\]

\[\sqrt{x} = (x - 2)^{2}\]

\[x = 1.\]

\[\textbf{а)}\ \sqrt{x} > 0\]

\[\sqrt{x} \neq 0\]

\[x \neq 0.\]

\[\textbf{б)}\ (x - 2)^{2} > 0\]

\[x - 2 \neq 0\ \]

\[x \neq 2.\]

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

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

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

\[= \frac{2}{3} + \frac{1}{3} \bullet 0^{3} - \frac{1}{3} \bullet ( - 1)^{3} =\]

\[= \frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1.\]

\[Ответ:\ \ 1.\]

\[2)\ y = x^{3}\text{\ \ }и\ \ t = 2x - x^{2}\]

\[x^{3} = 2x - x^{2}\]

\[x^{3} + x^{2} - 2x = 0\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\ \ \]

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

\[(x + 2) \bullet x \bullet (x - 1) = 0\]

\[x_{1} = - 2,\ \ \ x_{2} = 0,\ \ \ x_{3} = 1.\]

\[\textbf{а)}\ x^{3} > 0\]

\[x > 0.\]

\[\textbf{б)}\ 2x - x^{2} > 0\]

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

\[0 < x < 2.\]

\[S =\]

\[= \int_{0}^{1}{x^{3}\text{\ dx}} + \int_{1}^{2}{\left( 2x - x^{2} \right)\text{\ dx}} =\]

\[= \left. \ \frac{x^{4}}{4}\ \right|_{0}^{1} + \left. \ \left( 2 \bullet \frac{x^{2}}{2} - \frac{x^{3}}{3} \right) \right|_{1}^{2} =\]

\[= \left. \ \frac{x^{4}}{4}\ \right|_{0}^{1} + \left. \ \left( x^{2} - \frac{x^{3}}{3} \right) \right|_{1}^{2} =\]

\[= \frac{1^{4}}{4} - \frac{0^{4}}{4} + 2^{2} - \frac{2^{3}}{3} - 1^{2} + \frac{1^{3}}{3} =\]

\[= \frac{1}{4} + 4 - \frac{8}{3} - 1 + \frac{1}{3} =\]

\[= 3 - \frac{7}{3} + \frac{1}{4} = \frac{36 - 28 + 3}{12} = \frac{11}{12}.\]

\[Ответ:\ \ \frac{11}{12}.\]