ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1010

\[\boxed{\mathbf{1010}\mathbf{.}}\]

\[1)\int_{1}^{2}\frac{3}{2x - 1} =\]

\[= \left. \ 3 \bullet \frac{1}{2} \bullet \ln(2x - 1) \right|_{1}^{2} =\]

\[= \frac{3}{2} \bullet \ln(2 \bullet 2 - 1) - \frac{3}{2} \bullet \ln(2 - 1) =\]

\[= \frac{3}{2} \bullet \left( \ln(4 - 1) - \ln 1 \right) =\]

\[= \frac{3}{2} \bullet \left( \ln 3 - \ln e^{0} \right) = \frac{3}{2}\ln 3;\]

\[2)\int_{0}^{1}{\frac{4}{3x + 2}\text{\ dx}} =\]

\[= \left. \ 4 \bullet \frac{1}{3} \bullet \ln(3x + 2) \right|_{0}^{1} =\]

\[= \frac{4}{3} \bullet \ln(3 + 2) - \frac{4}{3} \bullet \ln(0 + 2) =\]

\[= \frac{4}{3} \bullet \left( \ln 5 - \ln 2 \right) = \frac{4}{3}\ln\frac{5}{2};\]

\[3)\int_{0}^{\frac{\pi}{2}}{\sin\left( 2x + \frac{\pi}{3} \right)\text{\ dx}} =\]

\[= \left. \ - \frac{1}{2} \bullet \cos\left( 2x + \frac{\pi}{3} \right) \right|_{0}^{\frac{\pi}{2}} =\]

\[= \frac{1}{2}\cos\frac{\pi}{3} + \frac{1}{2}\cos\frac{\pi}{3} = \cos\frac{\pi}{3} = \frac{1}{2}\text{.\ }\]