ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1136

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

\[= - \cos\pi + \cos\frac{\pi}{2} =\]

\[= - ( - 1) + 0 = 1;\]

\[2)\ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{\cos x\text{\ dx}} = \left. \ \sin x \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}} =\]

\[= \sin\frac{\pi}{3} - \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{1}{2} =\]

\[= \frac{\sqrt{3} - 1}{2};\]

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

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

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

\[= \frac{1}{3} + 1 + 3 + \frac{8}{3} - 4 + 6 =\]

\[= \frac{9}{3} + 6 = 3 + 6 = 9;\]

\[4)\ \int_{1}^{2}{\left( x^{2} - 6x + 8 \right)\text{dx}} =\]

\[= \left. \ \left( \frac{x^{3}}{3} - 6 \bullet \frac{x^{2}}{2} + 8x \right) \right|_{1}^{2} =\]

\[= \left. \ \left( \frac{x^{3}}{3} - 3x^{2} + 8x \right) \right|_{1}^{2} =\]

\[= \frac{8}{3} - 12 + 16 - \frac{1}{3} + 3 - 8 =\]

\[= \frac{7}{3} - 1 = 2\frac{1}{3} - 1 = 1\frac{1}{3};\]

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

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

\[= \left( - \frac{1}{3} + 3 \right) - \left( - \frac{1}{1} + 1 \right) =\]

\[= - \frac{1}{3} + 3 = 2\frac{2}{3};\]

\[6)\ \int_{- 1}^{1}{\frac{2}{5 - 4x}\text{dx}} =\]

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

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

\[= - \frac{1}{2}\ln|5 - 4| + \frac{1}{2}\ln|5 + 4| =\]

\[= - \frac{1}{2}\ln 1 + \frac{1}{2}\ln 9 = \ln 9^{\frac{1}{2}} = \ln 3.\]