Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1025

\[\boxed{\mathbf{1025}\mathbf{.}}\]

\[1)\ v(t) = 3t^{2} + 1\ \ \]

\[t_{1} = 0;\text{\ \ }t_{2} = 4:\]

\[S = \int_{0}^{4}{\left( 3t^{2} + 1 \right)\text{\ dx}} =\]

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

\[S = 4^{3} + 4 - 0^{3} - 0 = 64 + 4 =\]

\[= 68\ (м).\]

\[Ответ:\ \ 68\ метров.\]

\[2)\ v(t) = 2t^{2} + t\ \ \ \]

\[t_{1} = 1;\text{\ \ }t_{2} = 3:\]

\[S = \int_{1}^{3}{\left( 2t^{2} + t \right)\text{\ dx}} =\]

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

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

\[S = 18 + \frac{9}{2} - \frac{2}{3} - \frac{1}{2} =\]

\[= 18 + 4 - \frac{2}{3} = 22 - \frac{2}{3} =\]

\[= 21\frac{1}{3}\ (м).\]

\[Ответ:\ \ 21\frac{1}{3}\ метров.\]