ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 372

Авторы:
Год:2020-2021-2022-2023
Тип:учебник

Задание 372

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

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

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

\[= \frac{4^{2}}{2} - 2 \bullet 4^{\frac{3}{2}} = \frac{16}{2} - 2 \bullet 2^{3} =\]

\[= 8 - 2 \bullet 8 = 8 - 16 = - 8.\]

\[2)\ \int_{1}^{9}{\left( 2x - \frac{3}{\sqrt{x}} \right)\text{dx}} =\]

\[= \left. \ \left( 2 \bullet \frac{x^{2}}{2} - 3 \bullet 2x^{\frac{1}{2}} \right) \right|_{1}^{9} =\]

\[= \left. \ \left( x^{2} - 6\sqrt{x} \right) \right|_{1}^{9} =\]

\[= \left( 9^{2} - 6\sqrt{9} \right) - \left( 1^{2} - 6\sqrt{1} \right) =\]

\[= 81 - 18 - 1 + 6 = 68.\]

\[3)\ \int_{- 1}^{2}{\frac{5x - 2}{\sqrt[3]{x}}\text{dx}} =\]

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

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

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

\[= 3\sqrt[3]{4} + 6.\]

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

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

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

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

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

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

\[= 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}.\]

\[5)\ \int_{1}^{4}{\sqrt{x}\left( 3 - \frac{7}{x} \right)\text{dx}} =\]

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

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

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

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

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

\[6)\ \int_{1}^{8}{4\sqrt[3]{x}\left( 1 - \frac{4}{x} \right)\text{dx}} =\]

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

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

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

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

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

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам