Страница 304 номер 1008, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить интеграл (1004—1011). 1) \(\int\limits_{-2}^{1} x(x + 3)(2x - 1)\,dx\); 2) \(\int\limits_{-1}^{0} (x + 1)(x^2 - 2)\,dx\); 3) \(\int\limits_{1}^{2} \left(x + \frac{1}{x}\right)^2 dx\); 4) \(\int\limits_{-2}^{-1} \frac{4}{x^2}\left(1 - \frac{2}{x}\right)dx\).
1) \(\int_{-2}^{1}{x(x + 3)(2x - 1)dx} = \int_{-2}^{1}{x\left( 2x^{2} + 5x - 3 \right)dx} =\)
\(= \int_{-2}^{1}{\left( 2x^{3} + 5x^{2} - 3x \right)dx} = \left. \ \left( \dfrac{1}{2}x^{4} + \dfrac{5}{3}x^{3} - \dfrac{3}{2}x^{2} \right) \right|_{-2}^{1} =\)
\(= \dfrac{1}{2} + \dfrac{5}{3} - \dfrac{3}{2} - 8 + \dfrac{40}{3} + 6 = - 1 + 15 - 8 + 6 = 12;\)
2) \(\int_{-1}^{0}{(x + 1)\left( x^{2} - 2 \right)dx} = \int_{-1}^{0}{\left( x^{3} - 2x + x^{2} - 2 \right)dx} =\)
\(= \left. \ \left( \dfrac{x^{4}}{4} - x^{2} + \dfrac{x^{3}}{3} - 2x \right) \right|_{-1}^{0} = - \dfrac{1}{4} + 1 + \dfrac{1}{3} - 2 = \dfrac{1}{12} - 1 = - \dfrac{11}{12};\)
3) \(\int_{1}^{2}{\left( x + \frac{1}{x} \right)^{2}dx} = \int_{1}^{2}{\left( x^{2} + 2 + \frac{1}{x^{2}} \right)dx} = \left. \ \left( \frac{x^{3}}{3} + 2x - \frac{1}{x} \right) \right|_{1}^{2} =\)
\(= \dfrac{8}{3} + 4 - \dfrac{1}{2} - \dfrac{1}{3} - 2 + 1 = \dfrac{11}{6} + 3 = 4\dfrac{5}{6};\)
4) \(\int_{-2}^{-1}{\frac{4}{x^{2}}\left( 1 - \frac{2}{x} \right)dx} = \int_{-2}^{-1}{\left( \frac{4}{x^{2}} - \frac{8}{x^{3}} \right)dx} = \left. \ \left(-\frac{4}{x} + \frac{4}{x^{2}} \right) \right|_{-2}^{-1} =\)
\(= 4 + 4 - 2 - 1 = 5;\)
