Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1034

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

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

\[= 4 + 2 = 6;\]

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

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

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

\[= 3 \bullet 2 - \frac{2^{2}}{2} - 3 \bullet ( - 2) + \frac{( - 2)^{2}}{2} =\]

\[= 6 - 2 + 6 + 2 = 12;\]

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

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

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

\[= \frac{3^{3}}{3} - 3^{2} - \frac{1^{3}}{3} + 1^{2} =\]

\[= 9 - 9 - \frac{1}{3} + 1 = 1 - \frac{1}{3} = \frac{2}{3};\]

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

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

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

\[= 1^{2} - 1^{3} - ( - 1)^{2} + ( - 1)^{3} =\]

\[= 1 - 1 - 1 - 1 = - 2;\]

\[5)\ \int_{1}^{8}{\sqrt[3]{x}\text{\ dx}} = \int_{1}^{8}{x^{\frac{1}{3}}\text{\ dx}} =\]

\[= \left. \ \left( x^{\frac{4}{3}}\ :\frac{4}{3} \right) \right|_{1}^{8} = \left. \ \left( \frac{3}{4} \bullet x \bullet \sqrt[3]{x} \right) \right|_{1}^{8} =\]

\[= \frac{3}{4} \bullet 8 \bullet \sqrt[3]{8} - \frac{3}{4} \bullet 1 \bullet \sqrt[3]{1} =\]

\[= \frac{24}{4} \bullet 2 - \frac{3}{4} = \frac{45}{4} = 11\frac{1}{4};\]

\[6)\ \int_{1}^{2}{\frac{1}{x^{3}}\text{\ dx}} = \int_{1}^{2}{x^{- 3}\text{\ dx}} =\]

\[= \left. \ \frac{x^{- 2}}{- 2} \right|_{1}^{2} = \left. \ - \frac{1}{2x^{2}} \right|_{1}^{2} =\]

\[= - \frac{1}{2 \bullet 2^{2}} + \frac{1}{2 \bullet 1^{2}} = - \frac{1}{8} + \frac{1}{2} =\]

\[= - \frac{1}{8} + \frac{4}{8} = \frac{3}{8};\]

\[7)\ \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{\cos x\text{\ dx}} = \left. \ \sin x \right|_{- \frac{\pi}{2}}^{\frac{\pi}{2}} =\]

\[= \sin\frac{\pi}{2} - \sin\left( - \frac{\pi}{2} \right) =\]

\[= \sin\frac{\pi}{2} + \sin\frac{\pi}{2} = 2\sin\frac{\pi}{2} =\]

\[= 2 \bullet 1 = 2.\]