вправа 11.4 гдз 11 клас математика Мерзляк Номіровський 2019
Вправа 11.4
Умова:
Умова:
Обчисліть визначений інтеграл:
Відповідь ГДЗ:
\begin{equation} 1) \int_{-4}^{-2}2dx=2x \mid^{-2}_{-4} = \end{equation} \begin{equation} =2*(-2)-2*(-4)= \end{equation} \begin{equation} =-4+8=4; \end{equation} \begin{equation} 2) \int_{2}^{1} x^{3}dx=\frac{x^{4}}{4} \mid^{2}_{1}= \end{equation} \begin{equation} =\frac{2^{4}}{4}--\frac{1^{4}}{4}= \end{equation} \begin{equation} =\frac{16}{4}-\frac{1}{4}=\frac{15}{4}; \end{equation} \begin{equation} \end{equation} \begin{equation} 3) \int_{0}^{\frac{ \pi }{2}}\cos xdx= \sin x \mid^{\frac{ \pi }{2}}_{0}= \end{equation} \begin{equation} =\sin \frac{\pi }{2}- \sin 0=1; \end{equation} \begin{equation} 4)\int_{\frac{ \pi }{4}}^{\frac{ \pi }{2}}\frac{dx}{\sin^{2} x}= \end{equation} \begin{equation} =-ctg x \mid^{\frac{ \pi }{2}}_{\frac{ \pi }{4}}= \end{equation} \begin{equation} =-ctg\frac{\pi }{2}+ctg\frac{\pi }{4}= \end{equation} \begin{equation} =0+1=1; \end{equation} \begin{equation} 5) \int_{1}^{3}\frac{dx}{x^{4}}= \end{equation} \begin{equation} =-\frac{3}{x^{3}} \mid^{3}_{1} = \end{equation} \begin{equation} =-\frac{3}{3^{3}}+\frac{3}{1^{3}}= \end{equation} \begin{equation} =-\frac{3}{27}+3=3 \end{equation} \begin{equation} -\frac{1}{9}=2\frac{8}{9}; \end{equation} \begin{equation} 6) \int_{0}^{4}e^{x}dx=e^{x} \mid^{4}_{0} = \end{equation} \begin{equation} =e^{4}-e^{0}=e^{4}-1; \end{equation} \begin{equation} 7) \int_{1}^{e} \frac{dx}{x}= \ln \left | x \right | \mid^{e}_{1} = \end{equation} \begin{equation} \ln e- \ln 1=1-0=1. \end{equation}
\begin{equation} 1) \int_{-4}^{-2}2dx=2x \mid^{-2}_{-4} = \end{equation} \begin{equation} =2*(-2)-2*(-4)= \end{equation} \begin{equation} =-4+8=4; \end{equation} \begin{equation} 2) \int_{2}^{1} x^{3}dx=\frac{x^{4}}{4} \mid^{2}_{1}= \end{equation} \begin{equation} =\frac{2^{4}}{4}--\frac{1^{4}}{4}= \end{equation} \begin{equation} =\frac{16}{4}-\frac{1}{4}=\frac{15}{4}; \end{equation} \begin{equation} \end{equation} \begin{equation} 3) \int_{0}^{\frac{ \pi }{2}}\cos xdx= \sin x \mid^{\frac{ \pi }{2}}_{0}= \end{equation} \begin{equation} =\sin \frac{\pi }{2}- \sin 0=1; \end{equation} \begin{equation} 4)\int_{\frac{ \pi }{4}}^{\frac{ \pi }{2}}\frac{dx}{\sin^{2} x}= \end{equation} \begin{equation} =-ctg x \mid^{\frac{ \pi }{2}}_{\frac{ \pi }{4}}= \end{equation} \begin{equation} =-ctg\frac{\pi }{2}+ctg\frac{\pi }{4}= \end{equation} \begin{equation} =0+1=1; \end{equation} \begin{equation} 5) \int_{1}^{3}\frac{dx}{x^{4}}= \end{equation} \begin{equation} =-\frac{3}{x^{3}} \mid^{3}_{1} = \end{equation} \begin{equation} =-\frac{3}{3^{3}}+\frac{3}{1^{3}}= \end{equation} \begin{equation} =-\frac{3}{27}+3=3 \end{equation} \begin{equation} -\frac{1}{9}=2\frac{8}{9}; \end{equation} \begin{equation} 6) \int_{0}^{4}e^{x}dx=e^{x} \mid^{4}_{0} = \end{equation} \begin{equation} =e^{4}-e^{0}=e^{4}-1; \end{equation} \begin{equation} 7) \int_{1}^{e} \frac{dx}{x}= \ln \left | x \right | \mid^{e}_{1} = \end{equation} \begin{equation} \ln e- \ln 1=1-0=1. \end{equation}