Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 392

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

\[= 2 \bullet 2 - 2 \bullet ( - 1) = 4 + 2 = 6.\]

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

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

\[= \left( 6 - \frac{4}{2} \right) - \left( - 6 - \frac{4}{2} \right) = 12.\]

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

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

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

\[= (9 - 9) - \left( \frac{1}{3} - 1 \right) = \frac{2}{3}.\]

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

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

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

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

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

\[5)\ \int_{1}^{8}{\sqrt[3]{x}\text{\ dx}} = \left. \ \frac{3}{4}x^{\frac{4}{3}} \right|_{1}^{8} =\]

\[= \frac{3}{4} \bullet 8^{\frac{4}{3}} - \frac{3}{4} \bullet 1^{\frac{4}{3}} = \frac{3}{4} \bullet 2^{4} - \frac{3}{4} =\]

\[= \frac{3}{4} \bullet 16 - \frac{3}{4} = 12 - \frac{3}{4} = 11\frac{1}{4}.\]

\[6)\ \int_{1}^{2}{\frac{1}{x^{3}}\text{dx}} = \left. \ \frac{x^{- 2}}{- 2} \right|_{1}^{2} = \left. \ - \frac{1}{2x^{2}} \right|_{1}^{2} =\]

\[= - \frac{1}{2 \bullet 2^{2}} + \frac{1}{2 \bullet 1^{2}} = \frac{1}{2} - \frac{1}{8} =\]

\[= \frac{4 - 1}{8} = \frac{3}{8}.\]

\[7)\ \int_{0}^{\frac{\pi}{2}}{\sin x\text{dx}} = \left. \ - \cos x \right|_{0}^{\frac{\pi}{2}} =\]

\[= - \cos\frac{\pi}{2} + \cos 0 = 1.\]

\[8)\ \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{\cos x\text{dx}} = \left. \ \sin x \right|_{- \frac{\pi}{2}}^{\frac{\pi}{2}} =\]

\[= \sin\frac{\pi}{2} - \sin\left( - \frac{\pi}{2} \right) = 2.\]