ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1000

\[\boxed{\mathbf{1000}\mathbf{.}}\]

\[1)\ a = 2,\ \ \ b = 4:\]

\[\ f(x) = x^{3}\]

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

\[S = \int_{2}^{4}{x^{3}\text{\ dx}} = F(4) - F(2) =\]

\[= \frac{4^{4}}{4} - \frac{2^{4}}{4} = \frac{256 - 16}{4} = \frac{240}{4} =\]

\[= 60.\]

\[Ответ:\ \ 60.\]

\[2)\ a = 3,\ \ \ b = 4:\text{\ \ }\]

\[f(x) = x^{2}\]

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

\[S = \int_{3}^{4}{x^{2}\text{\ dx}} = F(4) - F(3) =\]

\[= \frac{4^{3}}{3} - \frac{3^{3}}{3} = \frac{64 - 27}{3} = \frac{37}{3} =\]

\[= 12\frac{1}{3}.\]

\[Ответ:\ \ 12\frac{1}{3}.\]

\[3)\ a = - 2,\ \ \ b = 1:\text{\ \ }\]

\[f(x) = x^{2} + 1\]

\[F(x) = \frac{x^{3}}{3} + \frac{x^{1}}{1} = \frac{x^{3}}{3} + x + C.\]

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

\[= F(1) - F( - 2);\]

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

\[= \frac{1}{3} + \frac{8}{3} + 3 = 3 + 3 = 6.\]

\[Ответ:\ \ 6.\]

\[4)\ a = 0,\ \ \ b = 2:\ \]

\[f(x) = x^{3} + 1\]

\[F(x) = \frac{x^{4}}{4} + \frac{x^{1}}{1} = \frac{x^{4}}{4} + x + C.\]

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

\[= F(2) - F(0);\]

\[S = \frac{2^{4}}{4} + 2 - \frac{0^{4}}{4} - 0 = \frac{16}{4} + 2 =\]

\[= 4 + 2 = 6.\]

\[Ответ:\ \ 6.\]

\[5)\ a = \frac{\pi}{3},\ \ \ b = \frac{2\pi}{3}:\]

\[f(x) = \sin x\]

\[F(x) = - \cos x + C.\]

\[S = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}{\sin x\text{\ dx}} =\]

\[= F\left( \frac{2\pi}{3} \right) - F\left( \frac{\pi}{3} \right);\]

\[S = - \cos\frac{2\pi}{3} + \cos\frac{\pi}{3} =\]

\[= - \left( - \frac{1}{2} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1.\]

\[Ответ:\ \ 1.\]

\[6)\ a = - \frac{\pi}{6},\ \ \ b = 0:\ \ \]

\[f(x) = \cos x\]

\[F(x) = \sin x + C.\]

\[S = \int_{- \frac{\pi}{6}}^{0}{\cos x\text{\ dx}} =\]

\[= F(0) - F\left( - \frac{\pi}{6} \right);\]

\[S = \sin 0 - \sin\left( - \frac{\pi}{6} \right) =\]

\[= 0 + \sin\frac{\pi}{6} = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]