Страница 304 номер 1011, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить интеграл (1004—1011). 1) \(\int\limits_{-\pi}^{\pi} \sin^2 x\,dx\); 2) \(\int\limits_{0}^{\frac{\pi}{2}} \sin x \cos x\,dx\); 3) \(\int\limits_{0}^{\frac{\pi}{4}} (\cos^2 x - \sin^2 x)\,dx\); 4) \(\int\limits_{0}^{\pi} (\sin^4 x + \cos^4 x)\,dx\); 5) \(\int\limits_{0}^{3} x^2\sqrt{x + 1}\,dx\); 6) \(\int\limits_{3}^{4} \frac{x^2 - 4x + 5}{x - 2}\,dx\).
1) \(\int_{-\pi}^{\pi}{\sin^{2}xdx} = \int_{-\pi}^{\pi}{\frac{1 - \cos{2x}}{2}dx} = \left. \ \left( \frac{1}{2}x - \frac{1}{4}\sin{2x} \right) \right|_{-\pi}^{\pi} = \frac{\pi}{2} + \frac{\pi}{2} = \pi\);
2) \(\int_{0}^{\frac{\pi}{2}}{\sin x\cos xdx} = \int_{0}^{\frac{\pi}{2}}{\frac{1}{2}\sin{2x}dx} = \left. \ - \frac{1}{4}\cos{2x} \right|_{0}^{\frac{\pi}{2}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\);
3) \(\int_{0}^{\frac{\pi}{4}}{\left( \cos^{2}x - \sin^{2}x \right)dx =}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx} = \left. \ \frac{1}{2}\sin{2x} \right|_{0}^{\frac{\pi}{4}} = \frac{1}{2} - 0 = \frac{1}{2}\);
4) \(\int_{0}^{\pi}{\left( \sin^{4}x + \cos^{4}x \right)dx} = \int_{0}^{\pi}{\left( 1 - \frac{1}{2}\sin^{2}{2x} \right)dx} = \int_{0}^{\pi}{\frac{4 - 1 + \cos{4x}}{4}dx} =\)
\(= \left. \ \left( \dfrac{3}{4}x + \dfrac{\sin{4x}}{16} \right) \right|_{0}^{\pi} = \dfrac{3\pi}{4} + 0 - 0 - 0 = \dfrac{3\pi}{4};\)
5) \(\int_{0}^{3}{x^{2}\sqrt{x + 1}dx} = \int_{0}^{3}{\left( x^{2} + 2x + 1 - 2x - 1 \right)\sqrt{x + 1}dx} =\)
\(= \int_{0}^{3}{\left( (x + 1)^{2} - 2x - 1 \right)\sqrt{x + 1}dx} =\)
\(= \int_{0}^{3}{(x + 1)^{\frac{5}{2}}dx} - 2\int_{0}^{3}{x\sqrt{x + 1}dx} - \int_{0}^{3}{(x + 1)^{\frac{1}{2}}dx} =\)
\(= \int_{0}^{3}{(x + 1)^{\frac{5}{2}}dx} - \int_{0}^{3}{(x + 1)^{\frac{1}{2}}dx} - 2\int_{0}^{3}{(x + 1 - 1)(x + 1)^{\frac{1}{2}}dx} =\)
\(= \int_{0}^{3}{(x + 1)^{\frac{5}{2}}dx} - \int_{0}^{3}{(x + 1)^{\frac{1}{2}}dx} -\)
\(- 2\int_{0}^{3}{\left( (x + 1)^{\frac{3}{2}} - (x + 1)^{\frac{1}{2}} \right)dx =}\)
\(= \left. \ \left( \dfrac{2}{7}(x + 1)^{\frac{7}{2}} - \dfrac{2}{3}(x + 1)^{\frac{3}{2}} - 2\left( \dfrac{2}{5}(x + 1)^{\frac{5}{2}} - \dfrac{2}{3}(x + 1)^{\frac{3}{2}} \right) \right) \right|_{0}^{3} =\)
\(= \dfrac{2}{7} \cdot 128 - \dfrac{2}{3} \cdot 8 - 2\left( \dfrac{2}{5} \cdot 32 - \dfrac{2}{3} \cdot 8 \right) - \dfrac{2}{7} + \dfrac{2}{3} + 2\left( \dfrac{2}{5} - \dfrac{2}{3} \right) =\)
\(= \dfrac{256}{7} - \dfrac{16}{3} - \dfrac{128}{5} + \dfrac{32}{3} - \dfrac{2}{7} + \dfrac{2}{3} - \dfrac{8}{15} = \dfrac{1696}{105} = 16\dfrac{16}{105};\)
6) \(\int_{3}^{4}{\frac{x^{2} - 4x + 5}{x - 2}dx} = \int_{3}^{4}{\left( (x - 2) + \frac{1}{x - 2} \right)dx} =\)
\(= \left. \ \left( \frac{x^{2}}{2} - 2x + \ln(x - 2) \right) \right|_{3}^{4} =\)
\(= 8 - 8 + \ln 2 - \dfrac{9}{2} + 6 - \ln 1 = \ln 2 + \dfrac{3}{2}.\)
