ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 393

\[1)\ \int_{0}^{1}{\left( 5x^{4} - 8x^{3} \right)\text{dx}} =\]

\[= \left. \ \left( 5 \bullet \frac{x^{5}}{5} - 8 \bullet \frac{x^{4}}{4} \right) \right|_{0}^{1} =\]

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

\[= \left( 1^{5} - 2 \bullet 1^{4} \right) - \left( 0^{5} - 2 \bullet 0^{4} \right) =\]

\[= 1 - 2 = - 1.\]

\[2)\ \int_{- 1}^{1}{\left( 6x^{3} - 5x \right)\text{dx}} =\]

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

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

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

\[= \left( \frac{3}{2} - \frac{5}{2} \right) - \left( \frac{3}{2} - \frac{5}{2} \right) = 0.\]

\[3)\ \int_{1}^{4}{\sqrt{x}\left( 3 - \frac{7}{x} \right)\text{dx}} =\]

\[= \int_{1}^{4}{\left( 3x^{\frac{1}{2}} - 7x^{- \frac{1}{2}} \right)\text{dx}} =\]

\[= \left. \ \left( 3x^{\frac{3}{2}}\ :\frac{3}{2} - 7x^{\frac{1}{2}}\ :\frac{1}{2} \right) \right|_{1}^{4} =\]

\[= \left. \ \left( 2x^{\frac{3}{2}} - 14x^{\frac{1}{2}} \right) \right|_{1}^{4} =\]

\[= \left( 2 \bullet 4^{\frac{3}{2}} - 14 \bullet 4^{\frac{1}{2}} \right) - \left( 2 \bullet 1^{\frac{3}{2}} - 14 \bullet 1^{\frac{1}{2}} \right) =\]

\[= \left( 2 \bullet 2^{3} - 14 \bullet 2 \right) - (2 - 14) =\]

\[= (16 - 28) + 12 = 0.\]

\[4)\ \int_{1}^{8}{4\sqrt[3]{x}\left( 1 - \frac{4}{x} \right)\text{dx}} =\]

\[= \int_{1}^{8}{\left( 4x^{\frac{1}{3}} - 16x^{- \frac{2}{3}} \right)\text{dx}} =\]

\[= \left. \ \left( 4x^{\frac{4}{3}}\ :\frac{4}{3} - 16x^{\frac{1}{3}}\ :\frac{1}{3} \right) \right|_{1}^{8} =\]

\[= \left. \ \left( 3x^{\frac{4}{3}} - 48x^{\frac{1}{3}} \right) \right|_{1}^{8} =\]

\[= \left( 3 \bullet 8^{\frac{4}{3}} - 48 \bullet 8^{\frac{1}{3}} \right) - \left( 3 \bullet 1^{\frac{4}{3}} - 48 \bullet 1^{\frac{1}{3}} \right) =\]

\[= \left( 3 \bullet 2^{4} - 48 \bullet 2 \right) - (3 - 48) =\]

\[= (48 - 96) + 45 = - 3.\]

\[5)\ \int_{0}^{3}{\sqrt{x + 1}\text{\ dx}} = \left. \ \frac{2}{3}(x + 1)^{\frac{3}{2}} \right|_{0}^{3} =\]

\[= \frac{2}{3}(3 + 1)^{\frac{3}{2}} - \frac{2}{3}(0 + 1)^{\frac{3}{2}} =\]

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

\[= \frac{16}{3} - \frac{2}{3} = \frac{14}{3} = 4\frac{2}{3}.\]

\[6)\ \int_{2}^{6}{\sqrt{2x - 3}\text{\ dx}} =\]

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

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

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

\[= \frac{1}{3} \bullet 9^{\frac{3}{2}} - \frac{1}{3} \bullet 1^{\frac{3}{2}} =\]

\[= \frac{1}{3} \bullet 3^{3} - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} =\]

\[= 9 - \frac{1}{3} = 8\frac{2}{3}.\]