ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 384

\[1)\ v(t) = 3t^{2} + 1;t_{1} = 0;t_{2} = 4:\]

\[x(t) = 3 \bullet \frac{t^{3}}{3} + t = t^{3} + t;\]

\[S = \left( 4^{3} + 4 \right) - \left( 0^{3} + 0 \right) =\]

\[= 64 + 4 = 68\ м.\]

\[Ответ:\ \ 68\ м.\]

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

\[x(t) = 2 \bullet \frac{t^{3}}{3} + \frac{t^{2}}{2};\]

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

\[= \frac{54}{3} + \frac{9}{2} - \frac{2}{3} - \frac{1}{2} = \frac{52}{3} + \frac{8}{2} =\]

\[= 17\frac{1}{3} + 4 = 21\frac{1}{3}.\]

\[Ответ:\ \ 21\frac{1}{3}\ м.\]

\[3)\ v(t) = 6t^{2} + 4;t_{1} = 2;t_{2} = 3:\]

\[x(t) = 6 \bullet \frac{t^{3}}{3} + 4t = 2t^{3} + 4t;\]

\[S =\]

\[= \left( 2 \bullet 3^{3} + 4 \bullet 3 \right) - \left( 2 \bullet 2^{3} + 4 \bullet 2 \right) =\]

\[= 54 + 12 - 16 - 8 = 42.\]

\[Ответ:\ \ 42\ м.\]

\[4)\ v(t) = t^{2} - t + 3;t_{1} = 0;t_{2} = 5:\]

\[x(t) = \frac{t^{3}}{3} - \frac{t^{2}}{2} + 3t;\]

\[S =\]

\[= \left( \frac{5^{3}}{3} - \frac{5^{2}}{2} + 3 \bullet 5 \right) - \left( \frac{0^{3}}{3} - \frac{0^{2}}{2} + 3 \bullet 0 \right) =\]

\[= \frac{125}{3} - \frac{25}{2} + 15 = \frac{250 - 75}{6} + 15 =\]

\[S = \frac{175}{6} + 15 = 29\frac{1}{6} + 15 = 44\frac{1}{6}.\]

\[Ответ:\ \ 44\frac{1}{6}\ м.\]