вправа 7.3 гдз 11 клас математика Нелін Долгова 2019
Вправа 7.3
Умова:
Умова:
Обчисліть інтеграл:
Відповідь ГДЗ:
\begin{equation} 1)\int_{-\Pi }^{2\Pi }sin\frac{x}{3}dx= \end{equation} \begin{equation} =-3cos\frac{x}{3}|_{-\Pi }^{2\Pi }= \end{equation} \begin{equation} =-3(cos\frac{2\Pi }{3}-cos(\frac{-\Pi }{3}))= \end{equation} \begin{equation} =-3(-0,5-0,5)=3; \end{equation} \begin{equation} 2)\int_{-2}^{2}\frac{dx}{\sqrt{2x+5}}= \end{equation} \begin{equation} =\frac{1}{2}\cdot 2\sqrt{2x+5}|_{-2}^{2}= \end{equation} \begin{equation} =\sqrt{2x+5}|_{-2}^{2}= \end{equation} \begin{equation} =\sqrt{2\cdot 2+5}- \end{equation} \begin{equation} -\sqrt{2\cdot (-2)+5}= \end{equation} \begin{equation} =\sqrt{9}-1=3-1=2; \end{equation} \begin{equation} 3)\int_{0}^{3\Pi }\frac{dx}{cos^{2}\frac{x}{9}}= \end{equation} \begin{equation} =9tg\frac{x}{9}|_{0}^{3\Pi }= \end{equation} \begin{equation} =9tg\frac{3\Pi }{9}-3tg0= \end{equation} \begin{equation} =9tg\frac{\Pi }{9}=9\sqrt{3}; \end{equation} \begin{equation} 4)\int_{-2}^{6}\frac{dx}{\sqrt{x+3}}= \end{equation} \begin{equation} =2\sqrt{x+3}|_{-2}^{6}= \end{equation} \begin{equation} =2(\sqrt{6+3}-\sqrt{-2+3})= \end{equation} \begin{equation} =2(3-1)=4; \end{equation} \begin{equation} 5)\int_{0}^{\frac{2\Pi }{3}}(sin\frac{x}{4}+ \end{equation} \begin{equation} +cos\frac{x}{4})^{2}dx= \end{equation} \begin{equation} =\int_{0}^{\frac{2\Pi }{3}}(sin^{2}\frac{x}{4}+2cos\frac{x}{4}+ \end{equation} \begin{equation} +cos\frac{2x}{4})dx= \end{equation} \begin{equation} =\int_{0}^{\frac{2\Pi }{3}}(1+sin\frac{x}{2})dx= \end{equation} \begin{equation} =(x+(-2cos\frac{x}{2}))|_{0}^{\frac{2\Pi }{3}}= \end{equation} \begin{equation} =(x-2cos\frac{x}{2})|_{0}^{\frac{2\Pi }{3}}= \end{equation} \begin{equation} =\frac{2\Pi }{3}-2cos\frac{2\Pi }{6}-(-2cos0)= \end{equation} \begin{equation} =\frac{2\Pi }{3}-2\cdot \frac{1}{2}+2= \end{equation} \begin{equation} =\frac{2\Pi }{3}+1; \end{equation} \begin{equation} 6)\int_{0}^{2}(1+2x)^{3}dx= \end{equation} \begin{equation} =\frac{1}{2}\frac{(1+2x)^{4}}{4}|_{0}^{2}= \end{equation} \begin{equation} =\frac{1}{8}(1+2\cdot 2)^{4}-\frac{1}{8}(1+0)^{4}= \end{equation} \begin{equation} =\frac{1}{8}\cdot 625-\frac{1}{8}= \end{equation} \begin{equation} =\frac{624}{8}=78; \end{equation} \begin{equation} 7)\int_{0}^{\frac{\Pi }{12}}(1+cos2x)dx= \end{equation} \begin{equation} =(x+\frac{1}{2}sin2x)|_{0}^{\frac{\Pi }{12}}= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{2}sin2\cdot \end{equation} \begin{equation} \cdot \frac{\Pi }{12}-\frac{1}{2}sin0= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{2}sin\frac{\Pi }{6}= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{4}; \end{equation} \begin{equation} 8)\int_{1}^{4}(x+\frac{\sqrt{x}}{x})dx= \end{equation} \begin{equation} =\int_{1}^{4}(x+\frac{1}{\sqrt{x}})dx= \end{equation} \begin{equation} =(\frac{x^{2}}{2}+2\sqrt{x})|_{1}^{4}= \end{equation} \begin{equation} =\frac{4^{2}}{2}-2\sqrt{4}-(\frac{1}{2}+2)= \end{equation} \begin{equation} =8+4-2\frac{1}{2}=9,5. \end{equation} Відповідь: 1) 3; 2) 2; 3) 9√3; 4) 4; \begin{equation} 5)\frac{2\Pi }{3}+1; \end{equation} 6) 78; \begin{equation} 7)\frac{\Pi }{12}+\frac{1}{4}; \end{equation} 8) 9,5.
\begin{equation} 1)\int_{-\Pi }^{2\Pi }sin\frac{x}{3}dx= \end{equation} \begin{equation} =-3cos\frac{x}{3}|_{-\Pi }^{2\Pi }= \end{equation} \begin{equation} =-3(cos\frac{2\Pi }{3}-cos(\frac{-\Pi }{3}))= \end{equation} \begin{equation} =-3(-0,5-0,5)=3; \end{equation} \begin{equation} 2)\int_{-2}^{2}\frac{dx}{\sqrt{2x+5}}= \end{equation} \begin{equation} =\frac{1}{2}\cdot 2\sqrt{2x+5}|_{-2}^{2}= \end{equation} \begin{equation} =\sqrt{2x+5}|_{-2}^{2}= \end{equation} \begin{equation} =\sqrt{2\cdot 2+5}- \end{equation} \begin{equation} -\sqrt{2\cdot (-2)+5}= \end{equation} \begin{equation} =\sqrt{9}-1=3-1=2; \end{equation} \begin{equation} 3)\int_{0}^{3\Pi }\frac{dx}{cos^{2}\frac{x}{9}}= \end{equation} \begin{equation} =9tg\frac{x}{9}|_{0}^{3\Pi }= \end{equation} \begin{equation} =9tg\frac{3\Pi }{9}-3tg0= \end{equation} \begin{equation} =9tg\frac{\Pi }{9}=9\sqrt{3}; \end{equation} \begin{equation} 4)\int_{-2}^{6}\frac{dx}{\sqrt{x+3}}= \end{equation} \begin{equation} =2\sqrt{x+3}|_{-2}^{6}= \end{equation} \begin{equation} =2(\sqrt{6+3}-\sqrt{-2+3})= \end{equation} \begin{equation} =2(3-1)=4; \end{equation} \begin{equation} 5)\int_{0}^{\frac{2\Pi }{3}}(sin\frac{x}{4}+ \end{equation} \begin{equation} +cos\frac{x}{4})^{2}dx= \end{equation} \begin{equation} =\int_{0}^{\frac{2\Pi }{3}}(sin^{2}\frac{x}{4}+2cos\frac{x}{4}+ \end{equation} \begin{equation} +cos\frac{2x}{4})dx= \end{equation} \begin{equation} =\int_{0}^{\frac{2\Pi }{3}}(1+sin\frac{x}{2})dx= \end{equation} \begin{equation} =(x+(-2cos\frac{x}{2}))|_{0}^{\frac{2\Pi }{3}}= \end{equation} \begin{equation} =(x-2cos\frac{x}{2})|_{0}^{\frac{2\Pi }{3}}= \end{equation} \begin{equation} =\frac{2\Pi }{3}-2cos\frac{2\Pi }{6}-(-2cos0)= \end{equation} \begin{equation} =\frac{2\Pi }{3}-2\cdot \frac{1}{2}+2= \end{equation} \begin{equation} =\frac{2\Pi }{3}+1; \end{equation} \begin{equation} 6)\int_{0}^{2}(1+2x)^{3}dx= \end{equation} \begin{equation} =\frac{1}{2}\frac{(1+2x)^{4}}{4}|_{0}^{2}= \end{equation} \begin{equation} =\frac{1}{8}(1+2\cdot 2)^{4}-\frac{1}{8}(1+0)^{4}= \end{equation} \begin{equation} =\frac{1}{8}\cdot 625-\frac{1}{8}= \end{equation} \begin{equation} =\frac{624}{8}=78; \end{equation} \begin{equation} 7)\int_{0}^{\frac{\Pi }{12}}(1+cos2x)dx= \end{equation} \begin{equation} =(x+\frac{1}{2}sin2x)|_{0}^{\frac{\Pi }{12}}= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{2}sin2\cdot \end{equation} \begin{equation} \cdot \frac{\Pi }{12}-\frac{1}{2}sin0= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{2}sin\frac{\Pi }{6}= \end{equation} \begin{equation} =\frac{\Pi }{12}+\frac{1}{4}; \end{equation} \begin{equation} 8)\int_{1}^{4}(x+\frac{\sqrt{x}}{x})dx= \end{equation} \begin{equation} =\int_{1}^{4}(x+\frac{1}{\sqrt{x}})dx= \end{equation} \begin{equation} =(\frac{x^{2}}{2}+2\sqrt{x})|_{1}^{4}= \end{equation} \begin{equation} =\frac{4^{2}}{2}-2\sqrt{4}-(\frac{1}{2}+2)= \end{equation} \begin{equation} =8+4-2\frac{1}{2}=9,5. \end{equation} Відповідь: 1) 3; 2) 2; 3) 9√3; 4) 4; \begin{equation} 5)\frac{2\Pi }{3}+1; \end{equation} 6) 78; \begin{equation} 7)\frac{\Pi }{12}+\frac{1}{4}; \end{equation} 8) 9,5.