ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1036

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

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

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

\[= 1^{5} - 2 \bullet 1^{4} - 0^{5} + 2 \bullet 0^{4} =\]

\[= 1 - 2 = - 1;\]

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

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

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

\[= 24 - 10 - 1,5 + 2,5 = 15;\]

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

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

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

\[= 8 \bullet 2 - 14 \bullet 2 - 2 + 14 =\]

\[= 16 - 28 + 12 = 0;\]

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

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

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

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

\[= 24 \bullet 2 - 48 \bullet 2 - 3 + 48 =\]

\[= 48 - 96 + 45 = - 3;\]

\[5)\ \int_{0}^{3}{\sqrt{x + 1}\text{\ dx}} =\]

\[= \int_{0}^{3}{(x + 1)^{\frac{1}{2}}\text{\ dx}} =\]

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

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

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

\[= \frac{2}{3} \bullet \sqrt{64} - \frac{2}{3} \bullet \sqrt{1} = \frac{2}{3} \bullet 8 - \frac{2}{3} =\]

\[= \frac{14}{3} = 4\frac{2}{3};\]

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

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

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

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

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

\[= \frac{1}{3} \bullet \sqrt{729} - \frac{1}{3} = \frac{1}{3} \bullet 27 - \frac{1}{3} =\]

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