Упр.11.8 ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)
1) ?(1; 3)(4x^3-4x+3)dx; 2) ?(?/2; 3?/2)cos(x/3)dx; 3) ?(?/6; ?/3)3dx/sin^2(2x);
4) ?(-2; 1)(x-3)^2dx; 7) ?(-1; 1)dx/(3-2x); 10) ?(-6; 0)e^(-x/6)dx;
5) ?(1/5; 1)(5x-3)^5dx; 8) ?(0; 2?)(sin(x/6)+cos(5x))dx; 11) ?(-1; -1/2)dx/(4x+1)^3;
6) ?(2; 6)dx/v(3x-2); 9) ?(0; 2?)sin(?/3-3x)dx; 12) ?(12; 116)(x/4-2)^(1/3)dx.
$$\int\limits_{1}^{3}(4x^3-4x+3)\,dx=\left(x^4-2x^2+3x\right)\Big|_{1}^{3}$$
$$=(81-18+9)-(1-2+3)=72-2=70.$$$$\int\limits_{\pi/2}^{3\pi/2}\cos\frac{x}{3}\,dx=3\sin\frac{x}{3}\Big|_{\pi/2}^{3\pi/2}$$
$$=3\left(\sin\frac{\pi}{2}-\sin\frac{\pi}{6}\right)=3\left(1-\frac12\right)=\frac32.$$$$\int\limits_{\pi/6}^{\pi/3}\frac{3\,dx}{\sin^2 2x}=-\frac32\ctg 2x\Big|_{\pi/6}^{\pi/3}$$
$$=-\frac32\ctg\frac{2\pi}{3}+\frac32\ctg\frac{\pi}{3}$$
$$=\frac32\left(\ctg\frac{\pi}{3}-\ctg\frac{2\pi}{3}\right)=\frac32\left(\frac{\sqrt3}{3}+\frac{\sqrt3}{3}\right)=\sqrt3.$$$$\int\limits_{-2}^{1}(x-3)^2\,dx=\frac{(x-3)^3}{3}\Big|_{-2}^{1}$$
$$=\frac{(-2)^3}{3}-\frac{(-5)^3}{3}=-\frac83+\frac{125}{3}=39.$$$$\int\limits_{1/5}^{1}(5x-3)^5\,dx=\frac{(5x-3)^6}{30}\Big|_{1/5}^{1}$$
$$=\frac{2^6}{30}-\frac{(-2)^6}{30}=0.$$$$\int\limits_{2}^{6}\frac{dx}{\sqrt{3x-2}}=\frac{2}{3}\sqrt{3x-2}\Big|_{2}^{6}$$
$$=\frac{2}{3}(4-2)=\frac{4}{3}.$$$$\int\limits_{-1}^{1}\frac{dx}{3-2x}=-\frac12\ln|3-2x|\Big|_{-1}^{1}$$
$$=-\frac12\ln 1+\frac12\ln 5=\frac12\ln 5.$$$$\int\limits_{0}^{2\pi}\left(\sin\frac{x}{6}+\cos 5x\right)\,dx=-6\cos\frac{x}{6}+\frac15\sin 5x\Big|_{0}^{2\pi}$$
$$=\left(-6\cos\frac{\pi}{3}+\frac15\sin 10\pi\right)-\left(-6\cos 0+\frac15\sin 0\right)=3.$$$$\int\limits_{0}^{2\pi}\sin\left(\frac{\pi}{3}-3x\right)\,dx=\frac13\cos\left(\frac{\pi}{3}-3x\right)\Big|_{0}^{2\pi}$$
$$=\frac13\left(\cos\left(\frac{\pi}{3}-6\pi\right)-\cos\frac{\pi}{3}\right)=0.$$$$\int\limits_{-6}^{0}e^{-x/6}\,dx=-6e^{-x/6}\Big|_{-6}^{0}$$
$$=-6e^0+6e^1=6e-6.$$$$\int\limits_{-1}^{-1/2}\frac{dx}{(4x+1)^3}=-\frac{1}{8(4x+1)^2}\Big|_{-1}^{-1/2}$$
$$=-\frac{1}{8\cdot 1}+\frac{1}{8\cdot 3^2}=-\frac18+\frac1{72}=-\frac19.$$$$\int\limits_{12}^{116}\left(\frac{x}{4}-2\right)^{1/3}dx=4\int\limits_{1}^{29}u^{1/3}\,du$$
$$=4\cdot\frac34u^{4/3}\Big|_{1}^{29}=3\left(29^{4/3}-1\right).$$
Ответ
$$1)\ 70;\quad 2)\ \frac32;\quad 3)\ \sqrt3;\quad 4)\ 39;\quad 5)\ 0;\quad 6)\ \frac43;\quad 7)\ \frac12\ln 5;\quad 8)\ 3;\quad 9)\ 0;\quad 10)\ 6e-6;\quad 11)\ -\frac19;\quad 12)\ 3\left(29^{4/3}-1\right).$$
