Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 994

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\[1)\ f(x) = \frac{2x^{4} - 4x^{3} + x}{3};\]

\[F(x) =\]

\[= \frac{1}{3} \bullet \left( 2 \bullet \frac{x^{5}}{5} - 4 \bullet \frac{x^{4}}{4} + \frac{x^{2}}{2} \right) =\]

\[= \frac{1}{3} \bullet \left( \frac{2}{5}x^{5} - x^{4} + \frac{1}{2}x^{2} \right) + C.\]

\[2)\ f(x) = \frac{6x^{3} - 3x + 2}{5};\]

\[F(x) =\]

\[= \frac{1}{5} \bullet \left( 6 \bullet \frac{x^{4}}{4} - 3 \bullet \frac{x^{2}}{2} + 2 \bullet \frac{x^{1}}{1} \right) =\]

\[= \frac{1}{5} \bullet \left( \frac{3}{2}x^{4} - \frac{3}{2}x^{2} + 2x \right) + C.\]

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

\[= x - 3 + 2x^{2} - 6x =\]

\[= 2x^{2} - 5x - 3;\]

\[F(x) = 2 \bullet \frac{x^{3}}{3} - 5 \bullet \frac{x^{2}}{2} - 3 \bullet \frac{x^{1}}{1} =\]

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

\[4)\ f(x) = (2x - 3)(2 + 3x) =\]

\[= 4x + 6x^{2} - 6 - 9x =\]

\[= 6x^{2} - 5x - 6;\]

\[F(x) = 6 \bullet \frac{x^{3}}{3} - 5 \bullet \frac{x^{2}}{2} - 6 \bullet \frac{x^{1}}{1} =\]

\[= 2x^{3} - \frac{5}{2}x^{2} - 6x + C.\]