Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 66

\[\boxed{\mathbf{66}.}\]

\[\textbf{а)}\ \int_{1}^{2}{(3x - 1)\text{dx}} =\]

\[= 3\int_{1}^{2}\text{xdx} - \int_{1}^{2}\text{dx};\]

\[\int_{1}^{2}\text{xdx} = \left. \ \frac{x^{2}}{2} \right|_{1}^{2} = \frac{2^{2}}{2} - \frac{1}{2} =\]

\[= 2 - 0,5 = 1,5;\]

\[\int_{1}^{2}\text{dx} = \left. \ x \right|_{1}^{2} = 2 - 1 = 1;\]

\[\int_{1}^{2}{(3x - 1)\text{dx}} = 3 \cdot 1,5 - 1 =\]

\[= 4,5 - 1 = 3,5.\]

\[\textbf{б)}\ \int_{- 2}^{3}{\left( x^{2} - 2x \right)\text{dx}} =\]

\[= \int_{- 2}^{3}{x^{2}\text{dx}} - 2\int_{- 2}^{3}\text{xdx};\]

\[\int_{- 2}^{3}{x^{2}\text{dx}} = \left. \ \frac{x^{3}}{3} \right|_{- 2}^{3} =\]

\[= \frac{3^{3}}{3} - \frac{( - 2)^{3}}{3} = \frac{27}{3} + \frac{8}{3} = \frac{35}{3};\]

\[\int_{- 2}^{3}\text{xdx} = \left. \ \frac{x^{2}}{2} \right|_{- 2}^{3} = \frac{3^{2}}{2} - \frac{( - 2)^{2}}{2} =\]

\[= \frac{9}{2} - \frac{4}{2} = \frac{5}{2};\]

\[\int_{- 2}^{3}{\left( x^{2} - 2x \right)\text{dx}} = \frac{35}{3} - 2 \cdot \frac{5}{2} =\]

\[= 11\frac{2}{3} - 5 = 6\frac{2}{3}.\]

\[\textbf{в)}\ \int_{0}^{2}{\left( 2x^{2} + 5x - 6 \right)\text{dx}} =\]

\[\int_{0}^{2}{x^{2}\text{dx}} = \left. \ \frac{x^{3}}{3} \right|_{0}^{2} = \frac{2^{3}}{3} - \frac{0^{3}}{3} = \frac{8}{3};\]

\[\int_{0}^{2}\text{xdx} = \left. \ \frac{x^{2}}{2} \right|_{0}^{2} = \frac{2^{2}}{2} - \frac{0}{2} = 2;\]

\[\int_{0}^{2}\text{dx} = \left. \ x \right|_{0}^{2} = 2 - 0 = 2;\]

\[\int_{0}^{2}{\left( 2x^{2} + 5x - 6 \right)\text{dx}} =\]

\[= 2 \cdot \frac{8}{3} + 5 \cdot 2 - 6 \cdot 2 =\]

\[= \frac{16}{3} + 10 - 12 = 5\frac{1}{3} - 2 = 3\frac{1}{3}.\]

\[\textbf{г)}\ \int_{- 2}^{1}{\left( - 2x^{2} - x + 8 \right)\text{dx}} =\]

\[\int_{- 2}^{1}{x^{2}\text{dx}} = \left. \ \frac{x^{3}}{3} \right|_{- 2}^{1} =\]

\[= \frac{1^{3}}{3} - \frac{( - 2)^{3}}{3} = \frac{1}{3} + \frac{8}{3} = 3;\]

\[\int_{- 2}^{1}\text{xdx} = \left. \ \frac{x^{2}}{2} \right|_{- 2}^{1} = \frac{1^{2}}{2} - \frac{( - 2)^{2}}{2} =\]

\[= \frac{1}{2} - \frac{4}{2} = - \frac{3}{2};\]

\[\ \int_{- 2}^{1}\text{dx} = \left. \ x \right|_{- 2}^{1} = 1 - ( - 2) = 3;\]

\[\int_{- 2}^{1}{\left( - 2x^{2} - x + 8 \right)\text{dx}} =\]

\[= - 2 \cdot 3 - 1 \cdot \left( - \frac{3}{2} \right) + 8 \cdot 3 =\]

\[= - 6 + 24 + 1,5 =\]

\[= 19,5.\]