ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1019

\[\boxed{\mathbf{1019}\mathbf{.}}\]

\[1)\ y = \sin x\ на\ отрезке\ \lbrack 0;\ \pi\rbrack\ и\ \]

\[точки\ (0;\ 0)\ и\ \left( \frac{\pi}{2};\ 1 \right);\]

\[\left\{ \begin{matrix} 0 = k \bullet 0 + b \\ 1 = k \bullet \frac{\pi}{2} + b \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} b = 0\ \ \ \ \ \ \\ 1 = k \bullet \frac{\pi}{2} \\ \end{matrix} \right.\ \]

\[\frac{\text{kπ}}{2} = 1 \rightarrow \ k = \frac{2}{\pi};\]

\[y = \frac{2}{\pi}x;\]

\[\textbf{а)}\ \sin x > 0\]

\[2\pi n < x < \pi + 2\pi n.\]

\[0 < x < \pi.\]

\[\textbf{б)}\ \frac{2}{\pi}x > 0\ \]

\[x > 0.\]

\[S = \int_{0}^{\frac{\pi}{2}}{\frac{2}{\pi}\text{x\ dx}} + \int_{\frac{\pi}{2}}^{\pi}{\sin x\text{dx}} =\]

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

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

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

\[= \frac{\pi}{4} + 1 + 0 = 1 + \frac{\pi}{4}.\]

\[Ответ:\ \ 1 + \frac{\pi}{4}.\]

\[2)\ y = \sin x;\ \ \ y = \cos x\ \]

\[на\ отрезке\ \left\lbrack 0;\ \frac{\pi}{2} \right\rbrack;\]

\[\sin x = \cos x\ \ \ \ \ |\ :\cos x\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n\]

\[x = \frac{\pi}{4} + \pi n\ \]

\[x = \frac{\pi}{4}.\]

\[\textbf{а)}\ \sin x > 0\]

\[2\pi n < x < \pi + 2\pi n\]

\[0 < x < \frac{\pi}{2}.\]

\[\textbf{б)}\ \cos x > 0\]

\[- \frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n\]

\[0 < x < \frac{\pi}{2}.\]

\[S = \int_{0}^{\frac{\pi}{4}}{\sin x\text{dx}} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\cos x\text{dx}} =\]

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

\[= - \frac{\sqrt{2}}{2} + 1 + 1 - \frac{\sqrt{2}}{2} = 2 - \sqrt{2}.\]

\[Ответ:\ \ 2 - \sqrt{2}.\]