Упр.11.9 ГДЗ Мерзляк 11 класс Базовый 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) ?(4?/3; 4?)sin(x/4)dx; 5) ?(4; 7)dx/v(3x+4); 8) ?(1; 7/6)dx/(6x-5)^2;
3) ?(0; ?)dx/cos^2(x/2-?/3); 6) ?(ln 3; ln 4)e^(-2x)dx; 9) ?(1; 4)v(7x-3)dx.

1. $$\int_{1}^{4}\left(\frac{4}{x^2}+2x-3x^2\right)\,dx =\left(-\frac{4}{x}+x^2-x^3\right)\Bigg|_{1}^{4}.$$
   Тогда
   $$\left(-\frac{4}{4}+4^2-4^3\right)-\left(-\frac{4}{1}+1^2-1^3\right) =(-1+16-64)-(-4+1-1)=-49+4=-45.$$

2. $$\int_{\frac{4\pi}{3}}^{4\pi}\sin\frac{x}{4}\,dx =-4\cos\frac{x}{4}\Bigg|_{\frac{4\pi}{3}}^{4\pi}.$$
   Тогда
   $$-4\cos\pi+4\cos\frac{\pi}{3} =-4(-1)+4\cdot\frac12=4+2=6.$$

3. $$\int_{0}^{\pi}\frac{dx}{\cos^2\left(\frac{x}{2}-\frac{\pi}{3}\right)} =\int_{0}^{\pi}\sec^2\left(\frac{x}{2}-\frac{\pi}{3}\right)\,dx.$$
   Положим $$u=\frac{x}{2}-\frac{\pi}{3},\quad du=\frac12\,dx,\quad dx=2\,du.$$
   Тогда
   $$\int 2\sec^2 u\,du=2\tan u,$$
   и
   $$2\tan\left(\frac{x}{2}-\frac{\pi}{3}\right)\Bigg|_{0}^{\pi} =2\tan\frac{\pi}{6}-2\tan\left(-\frac{\pi}{3}\right) =\frac{2\sqrt3}{3}+2\sqrt3=\frac{8\sqrt3}{3}.$$

4. $$\int_{0}^{1}(2x-1)^4\,dx.$$
   Подстановка $$u=2x-1,\quad du=2\,dx,\quad dx=\frac{du}{2}$$ даёт
   $$\int_{0}^{1}(2x-1)^4\,dx=\frac12\int_{-1}^{1}u^4\,du =\frac12\cdot\frac{u^5}{5}\Bigg|_{-1}^{1}.$$
   Тогда
   $$\frac{1}{10}\left(1^5-(-1)^5\right)=\frac{1}{10}(1+1)=\frac15.$$

5. $$\int_{4}^{7}\frac{dx}{\sqrt{3x+4}}.$$
   Подстановка $$u=3x+4,\quad du=3\,dx,\quad dx=\frac{du}{3}$$:
   $$\int_{4}^{7}\frac{dx}{\sqrt{3x+4}} =\frac13\int_{16}^{25}u^{-\frac12}\,du =\frac13\cdot 2\sqrt{u}\Bigg|_{16}^{25}.$$
   Получаем
   $$\frac23(5-4)=\frac23.$$

6. $$\int_{\ln 3}^{\ln 4}e^{-2x}\,dx =-\frac12 e^{-2x}\Bigg|_{\ln 3}^{\ln 4}.$$
   Тогда
   $$-\frac12 e^{-2\ln 4}+\frac12 e^{-2\ln 3} =-\frac12\cdot\frac{1}{16}+\frac12\cdot\frac{1}{9} =-\frac{1}{32}+\frac{1}{18} =\frac{7}{288}.$$

7. $$\int_{0}^{3}\frac{dx}{3x+1} =\frac13\ln|3x+1|\Bigg|_{0}^{3} =\frac13(\ln 10-\ln 1) =\frac13\ln 10.$$

8. $$\int_{1}^{\frac76}\frac{dx}{(6x-5)^2}.$$
   Подстановка $$u=6x-5,\quad du=6\,dx,\quad dx=\frac{du}{6}$$:
   $$\int_{1}^{\frac76}\frac{dx}{(6x-5)^2} =\frac16\int_{1}^{2}u^{-2}\,du =\frac16\left(-\frac1u\right)\Bigg|_{1}^{2}.$$
   Тогда
   $$-\frac16\cdot\frac12+\frac16\cdot 1=\frac{1}{12}.$$

9. $$\int_{1}^{4}\sqrt{7x-3}\,dx.$$
   Подстановка $$u=7x-3,\quad du=7\,dx,\quad dx=\frac{du}{7}$$:
   $$\int_{1}^{4}\sqrt{7x-3}\,dx =\frac17\int_{4}^{25}u^{\frac12}\,du =\frac17\cdot\frac{2}{3}u^{\frac32}\Bigg|_{4}^{25}.$$
   Получаем
   $$\frac{2}{21}\left(25^{\frac32}-4^{\frac32}\right) =\frac{2}{21}(125-8) =\frac{234}{21} =\frac{78}{7}.$$

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) $$\frac{78}{7}$$.