Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1002

\[\boxed{\mathbf{1002}\mathbf{.}}\]

\[1)\ a = 1,\ \ \ b = 8:\text{\ \ }\]

\[f(x) = \sqrt[3]{x} = x^{\frac{1}{3}};\]

\[F(x) = x^{\frac{4}{3}}\ :\frac{4}{3} = \frac{3}{4} \bullet x\sqrt[3]{x} + C.\]

\[S = \int_{1}^{8}{\sqrt[3]{x}\text{\ dx}} = F(8) - F(1);\]

\[S = \frac{3}{4} \bullet 8 \bullet \sqrt[3]{8} - \frac{3}{4} \bullet 1 \bullet \sqrt[3]{1} =\]

\[= \frac{24}{4} \bullet 2 - \frac{3}{4} = \frac{45}{4} = 11,25.\]

\[Ответ:\ \ 11,25.\]

\[2)\ a = 4,\ \ \ b = 9:\text{\ \ }\]

\[f(x) = \sqrt{x} = x^{\frac{1}{2}};\]

\[F(x) = x^{\frac{3}{2}}\ :\frac{3}{2} = \frac{2}{3} \bullet x\sqrt{x} + C.\]

\[S = \int_{1}^{8}{\sqrt{x}\text{\ dx}} = F(9) - F(4);\]

\[S = \frac{2}{3} \bullet 9\sqrt{9} - \frac{2}{3} \bullet 4\sqrt{4} =\]

\[= \frac{18}{3} \bullet 3 - \frac{8}{3} \bullet 2 = \frac{38}{3} = 12\frac{2}{3}.\]

\[Ответ:\ \ 12\frac{2}{3}.\]