Упр.11.9 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)
1) (1; 4)(4/x^2+2x-3x^2)dx; 4) (0; 1)(2x-1)^4dx; 7) (0; 3)dx/(3x+1);
2) (4п/3; 4п)sin(x/4)dx; 5) (4; 7)dx/v(3x+4); 8) (1; 7/6)dx/(6x-5)^2;
3) (0; п)dx/cos^2(x/2-п/3); 6) (ln 3; ln 4)e^(-2x)dx; 9) (1; 4)v(7x-3)dx.
$$\int_{1}^{4}\left(\frac{4}{x^2}+2x-3x^2\right)\,dx
=\left(-\frac{4}{x}+x^2-x^3\right)\Bigg|_{1}^{4}.$$$$\left(-\frac{4}{4}+4^2-4^3\right)-\left(-\frac{4}{1}+1^2-1^3\right)
=(-1+16-64)-(-4+1-1)=-49+4=-45.$$$$\int_{4\pi/3}^{4\pi}\sin\frac{x}{4}\,dx
=-4\cos\frac{x}{4}\Bigg|_{4\pi/3}^{4\pi}.$$$$-4\cos\pi+4\cos\frac{\pi}{3}
=-4(-1)+4\cdot\frac12=4+2=6.$$$$\int_{0}^{\pi}\frac{dx}{\cos^2\left(\frac{x}{2}-\frac{\pi}{3}\right)}
=2\tg\left(\frac{x}{2}-\frac{\pi}{3}\right)\Bigg|_{0}^{\pi}.$$$$2\tg\frac{\pi}{6}-2\tg\left(-\frac{\pi}{3}\right)
=2\cdot\frac{\sqrt3}{3}+2\cdot\sqrt3
=\frac{8\sqrt3}{3}.$$$$\int_{0}^{1}(2x-1)^4\,dx
=\frac12\cdot\frac{(2x-1)^5}{5}\Bigg|_{0}^{1}.$$$$\frac{1}{10}\left(1^5-(-1)^5\right)
=\frac{1}{10}\cdot(1+1)=\frac15.$$$$\int_{4}^{7}\frac{dx}{\sqrt{3x+4}}
=\frac13\cdot 2\sqrt{3x+4}\Bigg|_{4}^{7}.$$$$\frac23\left(\sqrt{25}-\sqrt{16}\right)
=\frac23(5-4)=\frac23.$$$$\int_{\ln 3}^{\ln 4}e^{-2x}\,dx
=-\frac12 e^{-2x}\Bigg|_{\ln 3}^{\ln 4}.$$$$-\frac12 e^{-2\ln 4}+\frac12 e^{-2\ln 3}
=-\frac12\cdot\frac{1}{16}+\frac12\cdot\frac{1}{9}
=\frac{1}{18}-\frac{1}{32}
=\frac{7}{288}.$$$$\int_{0}^{3}\frac{dx}{3x+1}
=\frac13\ln|3x+1|\Bigg|_{0}^{3}.$$$$\frac13\ln 10-\frac13\ln 1=\frac13\ln 10.$$
$$\int_{1}^{7/6}\frac{dx}{(6x-5)^2}
=-\frac16\cdot\frac{1}{6x-5}\Bigg|_{1}^{7/6}.$$$$-\frac16\left(\frac{1}{2}-1\right)=\frac{1}{12}.$$
$$\int_{1}^{4}\sqrt{7x-3}\,dx
=\frac{2}{21}(7x-3)^{3/2}\Bigg|_{1}^{4}.$$$$\frac{2}{21}\left(25^{3/2}-4^{3/2}\right)
=\frac{2}{21}(125-8)
=\frac{234}{21}
=\frac{78}{7}.$$
Ответ
1) $$-45$$; 2) $$6$$; 3) $$\frac{8\sqrt3}{3}$$; 4) $$\frac15$$; 5) $$\frac23$$; 6) $$\frac{7}{288}$$; 7) $$\frac13\ln 10$$; 8) $$\frac{1}{12}$$; 9) $$\frac{78}{7}$$.
