ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1003

\[\boxed{\mathbf{1003}\mathbf{.}}\]

\[1)\ b = 2:\text{\ \ }\]

\[f(x) = 5x - x^{2};\]

\[F(x) = 5 \bullet \frac{x^{2}}{2} - \frac{x^{3}}{3} + C.\]

\[Пересечения\ с\ осью\ x:\]

\[5x - x^{2} > 0\]

\[x \bullet (5 - x) > 0\]

\[x \bullet (x - 5) < 0\]

\[0 < x < 5.\]

\[S = \int_{2}^{5}{\left( 5x - x^{2} \right)\text{\ dx}} =\]

\[= F(5) - F(2);\]

\[S = 5 \bullet \frac{5^{2}}{2} - \frac{5^{3}}{3} - 5 \bullet \frac{2^{2}}{2} + \frac{2^{3}}{3} =\]

\[= \frac{125}{2} - \frac{125}{3} - \frac{20}{2} + \frac{8}{3} =\]

\[= 52,5 - 39 = 13,5.\]

\[Ответ:\ \ 13,5.\]

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

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

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

\[= \frac{x^{3}}{3} + x^{2} + C.\]

\[Пересечения\ с\ осью\ x:\]

\[x^{2} + 2x > 0\]

\[(x + 2) \bullet x > 0\]

\[x < - 2\ или\ x > 0.\]

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

\[= F(3) - F(0);\]

\[S = \frac{3^{3}}{3} + 3^{2} - \frac{0^{3}}{3} - 2 =\]

\[= 3^{2} + 3^{2} = 9 + 9 = 18.\]

\[Ответ:\ \ 18.\]

\[3)\ b = 1:\text{\ \ }\]

\[f(x) = e^{x} - 1;\]

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

\[Пересечения\ с\ осью\ x:\]

\[e^{x} - 1 > 0\]

\[e^{x} > 1\]

\[e^{x} > e^{0}\ \]

\[x > 0.\]

\[S = \int_{0}^{1}{\left( e^{x} - 1 \right)\text{\ dx}} =\]

\[= F(1) - F(0);\]

\[S = e^{1} - 1 - e^{0} + 0 =\]

\[= e - 1 - 1 = e - 2.\]

\[Ответ:\ \ e - 2.\]

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

\[f(x) = 1 - \frac{1}{x};\]

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

\[Пересечения\ с\ осью\ x:\]

\[1 - \frac{1}{x} > 0\]

\[x^{2} - x > 0\]

\[x \bullet (x - 1) > 0\]

\[x < 0\ или\ x > 1.\]

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

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

\[S = 2 - \ln 2 - 1 + \ln 1 =\]

\[= 1 - \ln 2 + \ln e^{0} = 1 - \ln 2.\]

\[Ответ:\ \ 1 - \ln 2.\]