Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1005

\[\boxed{\mathbf{1005}\mathbf{.}}\]

\[1)\int_{1}^{e}{\frac{1}{x}\text{\ dx}} = \left. \ \ln x \right|_{1}^{e} =\]

\[= \ln e - \ln 1 = \ln e^{1} - \ln e^{0} =\]

\[= 1 - 0 = 1;\]

\[2)\int_{0}^{\ln 2}{e^{x}\text{\ dx}} = \left. \ e^{x} \right|_{0}^{\ln 2} =\]

\[= e^{\ln 2} - e^{0} = 2 - 1 = 1;\]

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

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

\[= 0 + \sin\pi = 0;\]

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

\[= - \cos\pi + \cos( - 2\pi) =\]

\[= - ( - 1) + 1 = 2;\]

\[5)\int_{- 2\pi}^{\pi}{\sin{2x}\text{\ dx}} =\]

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

\[= - \frac{1}{2}\cos{2\pi} + \frac{1}{2}\cos( - 4\pi) =\]

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

\[6)\int_{- 3\pi}^{0}{\cos{3x}\text{\ dx}} =\]

\[= \left. \ \frac{1}{3}\sin{3x} \right|_{- 3\pi}^{0} =\]

\[= \frac{1}{3}\sin 0 - \frac{1}{3}\sin( - 9\pi) =\]

\[= \frac{1}{3} \bullet \sin\pi = 0.\ \]