Упр.11.8 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)

1) (1; 3)(4x^3-4x+3)dx; 2) (п/2; 3п/2)cos(x/3)dx; 3) (п/6; п/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; 2п)(sin(x/6)+cos(5x))dx; 11) (-1; -1/2)dx/(4x+1)^3;
6) (2; 6)dx/v(3x-2); 9) (0; 2п)sin(п/3-3x)dx; 12) (12; 116)(x/4-2)^(1/3)dx.

1. $$\int\limits_{1}^{3}(4x^3-4x+3)\,dx=\left(x^4-2x^2+3x\right)\Big|_{1}^{3}$$
   $$=(81-18+9)-(1-2+3)=72-2=70.$$
2. $$\int\limits_{\pi/2}^{3\pi/2}\cos\frac{x}{3}\,dx=3\sin\frac{x}{3}\Big|_{\pi/2}^{3\pi/2}$$
   $$=3\sin\frac{\pi}{2}-3\sin\frac{\pi}{6}=3-\frac{3}{2}=\frac{3}{2}.$$
3. $$\int\limits_{\pi/6}^{\pi/3}\frac{3\,dx}{\sin^2 2x}=-\frac{3}{2}\ctg 2x\Big|_{\pi/6}^{\pi/3}$$
   $$=-\frac{3}{2}\ctg\frac{2\pi}{3}+\frac{3}{2}\ctg\frac{\pi}{3}$$
   $$=-\frac{3}{2}\left(-\frac{\sqrt3}{3}\right)+\frac{3}{2}\cdot\frac{\sqrt3}{3}=\sqrt3.$$
4. $$\int\limits_{-2}^{1}(x-3)^2\,dx=\frac{(x-3)^3}{3}\Big|_{-2}^{1}$$
   $$=\frac{(1-3)^3}{3}-\frac{(-2-3)^3}{3}=-\frac{8}{3}+\frac{125}{3}=39.$$
5. $$\int\limits_{1/5}^{1}(5x-3)^5\,dx=\frac{1}{5}\cdot\frac{(5x-3)^6}{6}\Big|_{1/5}^{1}$$
   $$=\frac{1}{30}\left(2^6-(-2)^6\right)=0.$$
6. $$\int\limits_{2}^{6}\frac{dx}{\sqrt{3x-2}}=\frac{1}{3}\cdot 2\sqrt{3x-2}\Big|_{2}^{6}$$
   $$=\frac{2}{3}\left(\sqrt{16}-\sqrt{4}\right)=\frac{2}{3}(4-2)=\frac{4}{3}.$$
7. $$\int\limits_{-1}^{1}\frac{dx}{3-2x}=-\frac{1}{2}\ln|3-2x|\Big|_{-1}^{1}$$
   $$=-\frac{1}{2}\ln 1+\frac{1}{2}\ln 5=\frac{1}{2}\ln 5.$$
8. $$\int\limits_{0}^{2\pi}\left(\sin\frac{x}{6}+\cos 5x\right)\,dx=-6\cos\frac{x}{6}+\frac{1}{5}\sin 5x\Big|_{0}^{2\pi}$$
   $$=\left(-6\cos\frac{\pi}{3}+\frac{1}{5}\sin 10\pi\right)-\left(-6\cos 0+\frac{1}{5}\sin 0\right)$$
   $$=-3+6=3.$$
9. $$\int\limits_{0}^{2\pi}\sin\left(\frac{\pi}{3}-3x\right)\,dx=\frac{1}{3}\cos\left(\frac{\pi}{3}-3x\right)\Big|_{0}^{2\pi}$$
   $$=\frac{1}{3}\cos\left(\frac{\pi}{3}-6\pi\right)-\frac{1}{3}\cos\frac{\pi}{3}=0.$$
10. $$\int\limits_{-6}^{0}e^{-x/6}\,dx=-6e^{-x/6}\Big|_{-6}^{0}$$
    $$=-6e^0+6e^1=6e-6.$$
11. $$\int\limits_{-1}^{-1/2}\frac{dx}{(4x+1)^3}=-\frac{1}{2}\cdot\frac{1}{4}(4x+1)^{-2}\Big|_{-1}^{-1/2}$$
    $$=-\frac{1}{8}\left(\frac{1}{(-1)^2}-\frac{1}{(-3)^2}\right)=-\frac{1}{8}+\frac{1}{72}=-\frac{1}{9}.$$
12. $$\int\limits_{12}^{116}\left(\frac{x}{4}-2\right)^{1/3}dx=4\cdot\frac{3}{4}\left(\frac{x}{4}-2\right)^{4/3}\Big|_{12}^{116}$$
    $$=3\left(\sqrt[3]{27^4}-\sqrt[3]{1^4}\right)=3\cdot 81-3=240.$$

Ответ

1) $$70$$; 2) $$\frac{3}{2}$$; 3) $$\sqrt3$$; 4) $$39$$; 5) $$0$$; 6) $$\frac{4}{3}$$; 7) $$\frac{1}{2}\ln 5$$; 8) $$3$$; 9) $$0$$; 10) $$6e-6$$; 11) $$-\frac{1}{9}$$; 12) $$240$$.