Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 992

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\[1)\ f(x) = 2x + 3;\]

\[F(x) = 2 \bullet \frac{x^{2}}{2} + 3 \bullet \frac{x^{1}}{1} =\]

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

\[M(1;\ 2):\]

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

\[2 = 1 + 3 + C\]

\[2 = 4 + C\ \]

\[C = - 2.\]

\[Ответ:\ \ F(x) = x^{2} + 3x - 2.\]

\[2)\ f(x) = 4x - 1;\]

\[F(x) = 4 \bullet \frac{x^{2}}{2} - 1 \bullet \frac{x^{1}}{1} =\]

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

\[M( - 1;\ 3):\]

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

\[3 = 2 + 1 + C\]

\[3 = 3 + C\ \]

\[C = 0.\]

\[Ответ:\ \ F(x) = 2x^{2} - x.\]

\[3)\ f(x) = \sin{2x};\]

\[F(x) = - \frac{1}{2}\cos{2x} + C.\]

\[M\left( \frac{\pi}{2};\ 5 \right):\]

\[5 = - \frac{1}{2}\cos\pi + C\]

\[5 = - \frac{1}{2} \bullet ( - 1) + C\]

\[5 = \frac{1}{2} + C\]

\[C = \frac{9}{2}.\]

\[Ответ:\ \ F(x) = \frac{9 - \cos{2x}}{2}.\]

\[4)\ f(x) = \cos{3x};\]

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

\[M(0;\ 0):\]

\[0 = \frac{1}{3}\sin 0 + C\]

\[0 = \frac{1}{3} \bullet 0 + C\ \]

\[C = 0.\]

\[Ответ:\ \ F(x) = \frac{1}{3}\sin{3x}.\]