ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1011

\[\boxed{\mathbf{1011}\mathbf{.}}\]

\[1)\int_{- \pi}^{\pi}{\sin^{2}x\text{\ dx}} =\]

\[= \int_{- \pi}^{\pi}{\frac{1 - \cos{2x}}{2}\text{\ dx}} =\]

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

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

\[= \frac{\pi}{2} - \frac{\sin{2\pi}}{4} - \frac{- \pi}{2} + \frac{\sin( - 2\pi)}{2} =\]

\[= \frac{\pi}{2} - \frac{0}{4} + \frac{\pi}{2} + \frac{0}{2} = \pi;\]

\[2)\int_{0}^{\frac{\pi}{2}}{\sin x \bullet \cos x\text{\ dx}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{\left( \frac{1}{2}\sin{2x} \right)\text{\ dx}} =\]

\[= \left. \ \frac{1}{2} \bullet \left( - \frac{1}{2}\cos{2x} \right) \right|_{0}^{\frac{\pi}{2}} =\]

\[= \left. \ - \frac{\cos{2x}}{4} \right|_{0}^{\frac{\pi}{2}} =\]

\[= - \frac{\cos\left( 2 \bullet \frac{\pi}{2} \right)}{4} + \frac{\cos(2 \bullet 0)}{4} =\]

\[= - \frac{\cos\pi}{4} + \frac{\cos 0}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2};\]

\[3)\int_{0}^{\frac{\pi}{4}}{\left( \cos^{2}x - \sin^{2}x \right)\text{\ dx}} =\]

\[= \int_{0}^{\frac{\pi}{4}}{\cos{2x}\text{\ dx}} = \left. \ \frac{1}{2}\sin{2x} \right|_{0}^{\frac{\pi}{4}} =\]

\[= \frac{1}{2} \bullet \sin\left( 2 \bullet \frac{\pi}{4} \right) - \frac{1}{2} \bullet \sin(2 \bullet 0) =\]

\[= \frac{\sin\frac{\pi}{2}}{2} - \frac{\sin 0}{2} = \frac{1}{2} - \frac{0}{2} = \frac{1}{2};\]

\[4)\int_{0}^{\pi}{\left( \sin^{4}x + \cos^{4}x \right)\text{\ dx}} =\]

\[= \int_{0}^{\pi}{\left( 1 - \frac{1}{2}\sin^{2}{2x} \right)\text{\ dx}} =\]

\[= \int_{0}^{\pi}{\left( 1 - \frac{1 - \cos{4x}}{4} \right)\text{\ dx}} =\]

\[= \int_{0}^{\pi}{\left( \frac{3}{4} + \frac{\cos{4x}}{4} \right)\text{\ dx}} =\]

\[= \left. \ \left( \frac{3}{4} \bullet \frac{x^{1}}{1} + \frac{1}{4} \bullet \frac{1}{4}\sin{4x} \right) \right|_{0}^{\pi} =\]

\[= \left. \ \left( \frac{3x}{4} + \frac{\sin{4x}}{16} \right) \right|_{0}^{\pi} =\]

\[= \frac{3\pi}{4} + \frac{\sin{4\pi}}{16} - \frac{3 \bullet 0}{4} - \frac{\sin 0}{16} =\]

\[= \frac{3\pi}{4} + 0 - 0 - 0 = \frac{3\pi}{4};\]

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

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

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

\[= \frac{192 - 80 - 6 + 10}{15} = \frac{116}{15} =\]

\[= 7\frac{11}{15};\]

\[6)\int_{3}^{4}{\frac{x^{2} - 4x + 5}{x - 2}\text{\ dx}} =\]

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

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

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

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

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

\[= \frac{7}{2} - 2 + \ln 2 + \ln e^{0} =\]

\[3,5 - 2 + \ln 2 + 0 =\]

\[= 0,5 + \ln 2 = \frac{3}{2} + \ln 2.\]