ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1135

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

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

\[= \left. \ \left( (x - 1)^{\frac{4}{3}}\ :\frac{4}{3} \right) \right|_{2}^{9} =\]

\[= \left. \ \frac{3\sqrt[3]{(x - 1)^{4}}}{4} \right|_{2}^{9} =\]

\[= \frac{3\sqrt[3]{(9 - 1)^{4}}}{4} - \frac{3\sqrt[3]{(2 - 1)^{4}}}{4} =\]

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

\[= \frac{3 \bullet 16 - 3}{4} = \frac{3 \bullet 15}{4} =\]

\[= \frac{45}{4} = 11,25;\]

\[2)\ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\left( 2\cos^{2}x - 1 \right)\text{dx}} =\]

\[= \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\left( 2 \bullet \frac{1 + \cos{2x}}{2} - 1 \right)\text{dx}} =\]

\[= \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\cos{2x}\text{\ dx}} = \left. \ \frac{1}{2}\sin{2x} \right|_{\frac{\pi}{6}}^{\frac{\pi}{4}} =\]

\[= \frac{1}{2}\sin\frac{2\pi}{4} - \frac{1}{2}\sin\frac{2\pi}{6} =\]

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

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

\[= \frac{2 - \sqrt{3}}{4};\]

\[3)\ \int_{3}^{4}{\frac{x^{2} + 3}{x - 2}\text{dx}} =\]

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

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

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

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

\[= \frac{16}{2} + 8 + 7\ln 2 - \frac{9}{2} - 6 - 7\ln 1 =\]

\[= \frac{7}{2} + 2 + 7\ln 2 - 0 = 5,5 + 7\ln 2;\]

\[4)\ \int_{0}^{6}{x\sqrt{36 - x^{2}}\text{\ dx}} =\]

\[= \int_{0}^{6}{x\sqrt{u} \bullet \left( - \frac{1}{2x}\text{du} \right)} =\]

\[= \int_{0}^{6}{- \frac{\sqrt{u}}{2}\text{du}} = \int_{0}^{6}{- \frac{u^{\frac{1}{2}}}{2}\text{du}} =\]

\[= \left. \ \left( - \frac{1}{2} \bullet u^{\frac{3}{2}}\ :\frac{3}{2} \right) \right|_{0}^{6} = {- \left. \ \frac{1}{3}u^{\frac{3}{2}} \right|}_{0}^{6} =\]

\[= \left. \ - \frac{1}{3}\left( 36 - x^{2} \right)^{\frac{3}{2}} \right|_{0}^{6} =\]

\[= - \frac{1}{3}\left( 36 - 6^{2} \right)^{\frac{3}{2}} + \frac{1}{3}\left( 36 - 0^{2} \right)^{\frac{3}{2}} =\]

\[= \frac{1}{3} \bullet 36^{\frac{3}{2}} = \frac{1}{3} \bullet 6^{3} = \frac{216}{3} = 72.\]