Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1007

\[\boxed{\mathbf{1007}\mathbf{.}}\]

\[1)\int_{0}^{4}{\left( x - 3\sqrt{x} \right)\text{\ dx}} =\]

\[= \int_{0}^{4}{\left( x - 3x^{\frac{1}{2}} \right)\text{\ dx}} =\]

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

\[= \left. \ \left( \frac{x^{2}}{2} - 2x\sqrt{x} \right) \right|_{0}^{4} =\]

\[= \frac{4^{2}}{2} - 2 \bullet 4\sqrt{4} - \frac{0^{2}}{2} - 2 \bullet 0\sqrt{0} =\]

\[= \frac{16}{2} - 8 \bullet 2 = 8 - 16 = - 8;\]

\[2)\int_{1}^{9}{\left( 2x - \frac{3}{\sqrt{x}} \right)\text{\ dx}} =\]

\[= \int_{1}^{9}{\left( 2x - 3 \bullet x^{- \frac{1}{2}} \right)\text{\ dx}} =\]

\[= \left. \ \left( 2 \bullet \frac{x^{2}}{2} - 3 \bullet x^{\frac{1}{2}}\ :\frac{1}{2} \right) \right|_{1}^{9} =\]

\[= \left. \ \left( x^{2} - 6\sqrt{x} \right) \right|_{1}^{9} =\]

\[= 9^{2} - 6\sqrt{9} - 1^{2} + 6\sqrt{1} =\]

\[= 81 - 18 - 1 + 6 = 68;\]

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

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

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

\[4)\int_{1}^{3}{2e^{2x}\text{\ dx}} = \left. \ 2 \bullet \frac{1}{2}e^{2x} \right|_{1}^{3} =\]

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