Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 373

\[1)\ \int_{0}^{5}{\frac{6}{\sqrt{3x + 1}}\text{dx}} =\]

\[= \int_{0}^{5}{6(3x + 1)^{- \frac{1}{2}}\text{dx}} =\]

\[= \left. \ \left( 6 \bullet \frac{1}{3} \bullet (3x + 1)^{\frac{1}{2}} \bullet 2 \right) \right|_{0}^{5} =\]

\[= \left. \ 4\sqrt{3x + 1} \right|_{0}^{5} =\]

\[= 4\sqrt{3 \bullet 5 + 1} - 4\sqrt{3 \bullet 0 + 1} =\]

\[= 4\sqrt{16} - 4\sqrt{1} = 4 \bullet 4 - 4 \bullet 1 =\]

\[= 16 - 4 = 12.\]

\[2)\ \int_{2}^{7}{\frac{4}{\sqrt{x + 2}}\text{dx}} =\]

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

\[= \left. \ \left( 4 \bullet (x + 2)^{\frac{1}{2}} \bullet 2 \right) \right|_{2}^{7} = \left. \ 8\sqrt{x + 2} \right|_{2}^{7} =\]

\[= 8\sqrt{7 + 2} - 8\sqrt{2 + 2} =\]

\[= 8\sqrt{9} - 8\sqrt{4} = 8 \bullet 3 - 8 \bullet 2 =\]

\[= 24 - 16 = 8.\]