Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 390

\[1)\ f(x) = \cos x;\text{\ \ \ M}(0;\ - 2):\]

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

\[- 2 = \sin 0 + C\]

\[C = - 2.\]

\[Ответ:\ \ F(x) = \sin x - 2.\]

\[2)\ f(x) = \sin x;\text{\ \ \ M}( - \pi;\ 0):\]

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

\[0 = - \cos( - \pi) + C\]

\[C = - 1.\]

\[Ответ:\ \ F(x) = - \cos x - 1.\]

\[3)\ f(x) = \frac{1}{\sqrt{x}};\text{\ \ \ M}(4;\ 5):\]

\[F(x) = x^{\frac{1}{2}}\ :\frac{1}{2} + C = 2\sqrt{x} + C\]

\[5 = 2\sqrt{4} + C\]

\[C = 1.\]

\[Ответ:\ \ F(x) = 2\sqrt{x} + 1.\]

\[4)\ f(x) = e^{x};\text{\ \ \ M}(0;\ 2):\]

\[F(x) = e^{x} + C\]

\[2 = e^{0} + C\]

\[C = 1.\]

\[Ответ:\ \ F(x) = e^{x} + 1.\]

\[5)\ f(x) = 3x^{2} + 1;\text{\ M}(1;\ - 2):\]

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

\[= x^{3} + x + C\]

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

\[C = - 4.\]

\[Ответ:\ \ F(x) = x^{3} + x - 4.\]

\[6)\ f(x) = 2 - 2x;\text{\ M}(2;\ 3):\]

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

\[= 2x - x^{2} + C\]

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

\[C = 3.\]

\[Ответ:\ \ F(x) = 2x - x^{2} + 3.\]