Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1006

\[\boxed{\mathbf{1006}\mathbf{.}}\]

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

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

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

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

\[= 4 - 6 - 9 - 9 = - 20;\]

\[2)\int_{- 2}^{- 1}{(5 - 4x)\text{\ dx}} =\]

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

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

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

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

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

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

\[= 2 - 8 + 1 - 1 = - 6;\]

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

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

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

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

\[= \frac{1}{3} + 1 + \frac{1}{3} + 1 = 2\frac{2}{3};\]

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

\[= \left. \ \left( 3 \bullet \frac{x^{3}}{3} - 4 \bullet \frac{x^{2}}{2} + 5 \bullet \frac{x^{1}}{1} \right) \right|_{0}^{2} =\]

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