вправа 7.2 гдз 11 клас математика Нелін Долгова 2019
Вправа 7.2
Умова:
Умова:
Доведіть правильність рівності:
Відповідь ГДЗ:
\begin{equation} 1)\int_{0}^{\frac{\Pi }{4}}\frac{dx}{cos^{2}x}= \end{equation} \begin{equation} =tgx|_{\frac{\Pi }{4}}^{0}=tg\frac{\Pi }{4}=1 \end{equation} \begin{equation} \int_{0}^{1}dx=x|_{0}^{1}=1 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{2}}\frac{dx}{cos^{2}x}=\int_{0}^{1}dx; \end{equation} \begin{equation} 2)\int_{0}^{\frac{\Pi }{3}}sinxdx= \end{equation} \begin{equation} =-cosx|_{0}^{\frac{\Pi }{3}}= \end{equation} \begin{equation} =-cos\frac{\Pi }{3}+cos0= \end{equation} \begin{equation} =-\frac{1}{2}+1=0,5 \end{equation} \begin{equation} \int_{\frac{1}{16}}^{\frac{1}{4}}\frac{dx}{\sqrt{x}}= \end{equation} \begin{equation} =\int_{\frac{1}{16}}^{\frac{1}{4}}x^{-\frac{1}{2}}dx= \end{equation} \begin{equation} =\frac{x}{\frac{1}{2}}^{\frac{1}{2}}|_{\frac{1}{13}}^{\frac{1}{4}}= \end{equation} \begin{equation} =2\sqrt{x}|_{\frac{1}{16}}^{\frac{1}{4}}= \end{equation} \begin{equation} =2\cdot \sqrt{\frac{1}{4}}-2\sqrt{\frac{1}{16}}= \end{equation} \begin{equation} =2\cdot \frac{1}{2}-2\cdot \frac{1}{4}= \end{equation} \begin{equation} =1-\frac{1}{2}=0,5 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{3}}sinxdx=\int_{\frac{1}{16}}^{\frac{1}{4}}\frac{dx}{\sqrt{x}}; \end{equation} \begin{equation} 3)\int_{0}^{\frac{\Pi }{2}}cosxdx=sinx|_{0}^{\frac{\Pi }{2}}= \end{equation} \begin{equation} =sin\frac{\Pi }{2}-sin0=1 \end{equation} \begin{equation} \int_{0}^{3\sqrt{3}}x^{2}dx=\frac{x^{3}}{3}|_{0}^{3\sqrt{3}}= \end{equation} \begin{equation} =(\frac{\sqrt[3]{3}}{3})^{3}=1 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{2}}cosxdx= \end{equation} \begin{equation} \int_{0}^{3\sqrt{3}}x^{2}dx; \end{equation} \begin{equation} 4)\int_{0}^{1}(2x+1)dx= \end{equation} \begin{equation} =(\frac{2x^{3}}{3}+x)|_{0}^{1}= \end{equation} \begin{equation} =\frac{2}{3}+1=1\frac{2}{3} \end{equation} \begin{equation} \int_{0}^{2}(x^{3}-1)dx= \end{equation} \begin{equation} =(\frac{x^{4}}{4}-x)|_{0}^{2}= \end{equation} \begin{equation} =\frac{24}{4}-2=\frac{16}{4}-1= \end{equation} \begin{equation} =4-2=2. \end{equation}
\begin{equation} 1)\int_{0}^{\frac{\Pi }{4}}\frac{dx}{cos^{2}x}= \end{equation} \begin{equation} =tgx|_{\frac{\Pi }{4}}^{0}=tg\frac{\Pi }{4}=1 \end{equation} \begin{equation} \int_{0}^{1}dx=x|_{0}^{1}=1 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{2}}\frac{dx}{cos^{2}x}=\int_{0}^{1}dx; \end{equation} \begin{equation} 2)\int_{0}^{\frac{\Pi }{3}}sinxdx= \end{equation} \begin{equation} =-cosx|_{0}^{\frac{\Pi }{3}}= \end{equation} \begin{equation} =-cos\frac{\Pi }{3}+cos0= \end{equation} \begin{equation} =-\frac{1}{2}+1=0,5 \end{equation} \begin{equation} \int_{\frac{1}{16}}^{\frac{1}{4}}\frac{dx}{\sqrt{x}}= \end{equation} \begin{equation} =\int_{\frac{1}{16}}^{\frac{1}{4}}x^{-\frac{1}{2}}dx= \end{equation} \begin{equation} =\frac{x}{\frac{1}{2}}^{\frac{1}{2}}|_{\frac{1}{13}}^{\frac{1}{4}}= \end{equation} \begin{equation} =2\sqrt{x}|_{\frac{1}{16}}^{\frac{1}{4}}= \end{equation} \begin{equation} =2\cdot \sqrt{\frac{1}{4}}-2\sqrt{\frac{1}{16}}= \end{equation} \begin{equation} =2\cdot \frac{1}{2}-2\cdot \frac{1}{4}= \end{equation} \begin{equation} =1-\frac{1}{2}=0,5 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{3}}sinxdx=\int_{\frac{1}{16}}^{\frac{1}{4}}\frac{dx}{\sqrt{x}}; \end{equation} \begin{equation} 3)\int_{0}^{\frac{\Pi }{2}}cosxdx=sinx|_{0}^{\frac{\Pi }{2}}= \end{equation} \begin{equation} =sin\frac{\Pi }{2}-sin0=1 \end{equation} \begin{equation} \int_{0}^{3\sqrt{3}}x^{2}dx=\frac{x^{3}}{3}|_{0}^{3\sqrt{3}}= \end{equation} \begin{equation} =(\frac{\sqrt[3]{3}}{3})^{3}=1 \end{equation} Отже: \begin{equation} \int_{0}^{\frac{\Pi }{2}}cosxdx= \end{equation} \begin{equation} \int_{0}^{3\sqrt{3}}x^{2}dx; \end{equation} \begin{equation} 4)\int_{0}^{1}(2x+1)dx= \end{equation} \begin{equation} =(\frac{2x^{3}}{3}+x)|_{0}^{1}= \end{equation} \begin{equation} =\frac{2}{3}+1=1\frac{2}{3} \end{equation} \begin{equation} \int_{0}^{2}(x^{3}-1)dx= \end{equation} \begin{equation} =(\frac{x^{4}}{4}-x)|_{0}^{2}= \end{equation} \begin{equation} =\frac{24}{4}-2=\frac{16}{4}-1= \end{equation} \begin{equation} =4-2=2. \end{equation}