Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 374

\[1)\ \int_{0}^{2}{e^{3x}\text{\ dx}} = \left. \ \frac{1}{3}e^{3x} \right|_{0}^{2} =\]

\[= \frac{1}{3}e^{3 \bullet 2} - \frac{1}{3}e^{3 \bullet 0} = \frac{e^{6} - e^{0}}{3} =\]

\[= \frac{e^{6} - 1}{3}.\]

\[2)\ \int_{1}^{3}{2e^{2x}\text{\ dx}} = \left. \ \left( 2 \bullet \frac{1}{2}e^{2x} \right) \right|_{1}^{3} =\]

\[= \left. \ e^{2x} \right|_{1}^{3} = e^{2 \bullet 3} - e^{2 \bullet 1} = e^{6} - e^{2}.\]

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

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

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

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

\[4)\ \int_{- 1}^{1}{\frac{4}{3x + 5}\text{dx}} =\]

\[= \left. \ \left( 4 \bullet \frac{1}{3}\ln|3x + 5| \right) \right|_{- 1}^{1} =\]

\[= \frac{4}{3}\ln|3 \bullet 1 + 5| - \frac{4}{3}\ln\left| 3 \bullet ( - 1) + 5 \right| =\]

\[= \frac{4}{3}\left( \ln 8 - \ln 2 \right) = \frac{4}{3}\ln 4.\]