ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 369

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

\[= \left( 2^{2} - 3 \bullet 2 \right) - \left( ( - 3)^{2} - 3 \bullet ( - 3) \right) =\]

\[= (4 - 6) - (9 + 9) =\]

\[= - 2 - 18 = - 20.\]

\[2)\ \int_{- 2}^{- 1}{(5 - 4x)\text{dx}} = \left. \ \left( 5x - 2x^{2} \right) \right|_{- 2}^{- 1} =\]

\[= ( - 5 - 2) - ( - 10 - 8) =\]

\[= - 7 + 18 = 11;\]

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

\[= \left( 2 - 2^{3} \right) - \left( - 1 - ( - 1)^{3} \right) =\]

\[= (2 - 8) - ( - 1 + 1) = 2 - 8 = - 6.\]

\[4)\ \int_{- 1}^{1}{\left( x^{2} + 1 \right)\text{dx}} = \left. \ \left( \frac{x^{3}}{3} + x \right) \right|_{- 1}^{1} =\]

\[= \left( \frac{1^{3}}{3} + 1 \right) - \left( \frac{( - 1)^{3}}{3} - 1 \right) =\]

\[= \left( \frac{1}{3} + 1 \right) - \left( - \frac{1}{3} - 1 \right) = \frac{4}{3} + \frac{4}{3} =\]

\[= \frac{8}{3} = 2\frac{2}{3}.\]

\[5)\ \int_{1}^{2}{\left( 2x + 3x^{2} \right)\text{dx}} = \left. \ \left( x^{2} + x^{3} \right) \right|_{1}^{2} =\]

\[= \left( 2^{2} + 2^{3} \right) - \left( 1^{2} + 1^{3} \right) =\]

\[= (4 + 8) - (1 + 1) = 12 - 2 = 10.\]

\[6)\ \int_{- 2}^{0}{\left( 9x^{2} - 4x \right)\text{dx}} =\]

\[= \left. \ \left( 3x^{3} - 2x^{2} \right) \right|_{- 2}^{0} =\]

\[= - \left( 3 \bullet ( - 2)^{3} - 2 \bullet ( - 2)^{2} \right) =\]

\[= - ( - 24 - 8) = 32.\]

\[7)\ \int_{- 2}^{- 1}{\left( 6x^{2} + 2x - 10 \right)\text{dx}} =\]

\[= \left. \ \left( 2x^{3} + x^{2} - 10x \right) \right|_{- 2}^{- 1} =\]

\[= ( - 2 + 11) - ( - 16 + 24) =\]

\[= 9 - 8 = 1.\]

\[8)\ \int_{0}^{2}{\left( 3x^{2} - 4x + 5 \right)\text{dx}} =\]

\[= \left. \ \left( x^{3} - 2x^{2} + 5x \right) \right|_{0}^{2} =\]

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

\[= 8 - 8 + 10 = 10.\]