Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1004

\[\boxed{\mathbf{1004}\mathbf{.}}\]

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

\[2)\int_{0}^{3}{x^{2}\text{\ dx}} = \left. \ \frac{x^{3}}{3} \right|_{0}^{3} = \frac{3^{3}}{3} - \frac{0^{3}}{3} =\]

\[= 3^{2} = 9;\]

\[3)\int_{- 1}^{2}{3x^{2}\text{\ dx}} = \left. \ \frac{3x^{3}}{3} \right|_{- 1}^{2} =\]

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

\[= 8 + 1 = 9;\]

\[4)\int_{- 2}^{3}{2x\ dx} = \left. \ \frac{2x^{2}}{2} \right|_{- 2}^{3} = \left. \ x^{2} \right|_{- 2}^{3} =\]

\[= 3^{2} - ( - 2)^{2} = 9 - 4 = 5;\]

\[5)\int_{2}^{3}{\frac{1}{x^{2}}\text{\ dx}} = \int_{2}^{3}{x^{- 2}\text{\ dx}} =\]

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

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

\[6)\int_{1}^{2}{\frac{1}{x^{3}}\text{\ dx}} = \int_{1}^{2}{x^{- 3}\text{\ dx}} =\]

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

\[= - \frac{1}{2 \bullet 2^{2}} + \frac{1}{2 \bullet 1^{2}} = \frac{1}{2} - \frac{1}{8} = \frac{3}{8};\]

\[7)\int_{1}^{4}{\sqrt{x}\text{\ dx}} = \int_{1}^{4}{x^{\frac{1}{2}}\text{\ dx}} =\]

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

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

\[= \frac{8 \bullet 2}{3} - \frac{2}{3} = \frac{14}{3} = 4\frac{2}{3};\]

\[8)\int_{4}^{9}{\frac{1}{\sqrt{x}}\text{\ dx}} = \int_{4}^{9}{x^{- \frac{1}{2}}\text{\ dx}} =\]

\[= \left. \ \left( x^{\frac{1}{2}}\ :\frac{1}{2} \right) \right|_{4}^{9} = \left. \ 2\sqrt{x} \right|_{4}^{9} =\]

\[= 2\sqrt{9} - 2\sqrt{4} = 6 - 4 = 2.\]