Упр.28.403 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)
1) (1; 3)dx/x^2; 6) (-2; 2)dx/v(2x+5);
2) (0; п)(6cos(4x)-3sin(x))dx; 7) (0; 2)(3x-2)^3 dx;
3) (0; п/18)dx/sin^2(3x+п/6); 8) (2; 4)e^(-x) dx;
4) (-2; 1)(x^2-2x+4)dx; 9) (0; 5)dx/(4x+1).
5) (1; 3)(4/x-x)dx;
$$\int_{1}^{3}\frac{dx}{x^2}=\int_{1}^{3}x^{-2}\,dx=-x^{-1}\Big|_{1}^{3}=-\frac{1}{3}+1=\frac{2}{3}.$$
$$\int_{0}^{\pi}(6\cos 4x-3\sin x)\,dx=\left(\frac{6}{4}\sin 4x+3\cos x\right)\Big|_{0}^{\pi}.$$
$$=\left(\frac{3}{2}\sin 4\pi+3\cos\pi\right)-\left(\frac{3}{2}\sin 0+3\cos 0\right)=-3-3=-6.$$
$$\int_{0}^{\pi/18}\frac{dx}{\sin^2\left(3x+\frac{\pi}{6}\right)}=-\frac{1}{3}\ctg\left(3x+\frac{\pi}{6}\right)\Big|_{0}^{\pi/18}.$$
$$=-\frac{1}{3}\ctg\frac{\pi}{3}+\frac{1}{3}\ctg\frac{\pi}{6}=-\frac{1}{3}\cdot\frac{1}{\sqrt{3}}+\frac{1}{3}\cdot\sqrt{3}=\frac{2\sqrt{3}}{9}.$$
$$\int_{-2}^{1}(x^2-2x+4)\,dx=\left(\frac{x^3}{3}-x^2+4x\right)\Big|_{-2}^{1}.$$
$$=\left(\frac{1}{3}-1+4\right)-\left(-\frac{8}{3}-4-8\right)=\frac{10}{3}+\frac{44}{3}=18.$$
$$\int_{1}^{3}\left(\frac{4}{x}-x\right)\,dx=\left(4\ln|x|-\frac{x^2}{2}\right)\Big|_{1}^{3}.$$
$$=\left(4\ln 3-\frac{9}{2}\right)-\left(4\ln 1-\frac{1}{2}\right)=4\ln 3-4.$$
$$\int_{-2}^{2}\frac{dx}{\sqrt{2x+5}}.$$
Положим $$u=2x+5,$$ тогда $$du=2dx,$$ $$dx=\frac{du}{2}.$$ Получаем:
$$\int_{-2}^{2}\frac{dx}{\sqrt{2x+5}}=\frac{1}{2}\int_{1}^{9}u^{-1/2}\,du=\sqrt{u}\Big|_{1}^{9}=3-1=2.$$
$$\int_{0}^{2}(3x-2)^3\,dx.$$
Положим $$u=3x-2,$$ тогда $$du=3dx,$$ $$dx=\frac{du}{3}.$$ Тогда
$$\int_{0}^{2}(3x-2)^3\,dx=\frac{1}{3}\int_{-2}^{4}u^3\,du=\frac{1}{12}u^4\Big|_{-2}^{4}.$$
$$=\frac{4^4-(-2)^4}{12}=\frac{256-16}{12}=20.$$
$$\int_{2}^{4}e^{-x}\,dx=-e^{-x}\Big|_{2}^{4}=-e^{-4}+e^{-2}=\frac{e^2-1}{e^4}.$$
$$\int_{0}^{5}\frac{dx}{4x+1}=\frac{1}{4}\ln|4x+1|\Big|_{0}^{5}.$$
$$=\frac{1}{4}\ln 21-\frac{1}{4}\ln 1=\frac{1}{4}\ln 21.$$
Ответ
1) $$\frac{2}{3}$$; 2) $$-6$$; 3) $$\frac{2\sqrt{3}}{9}$$; 4) $$18$$; 5) $$4\ln 3-4$$; 6) $$2$$; 7) $$20$$; 8) $$\frac{e^2-1}{e^4}$$; 9) $$\frac{1}{4}\ln 21$$.
