Упр.21.2 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

1. а) $$\int_{1}^{2}\left(\frac{3}{x^2}+x^2+2\right)\,dx$$; в) $$\int_{-2}^{-1}\left(-\frac{5}{x^2}+x^4-3x\right)\,dx$$; б) $$\int_{0}^{1}\left(\frac{2}{\sqrt{x+1}}-\frac{3}{\sqrt{(x+1)^2}}\right)\,dx$$; г) $$\int_{5}^{8}\left(\frac{2}{(x-2)^2}-\frac{1}{\sqrt{x-4}}\right)\,dx$$.

Вычислим интегралы по формуле Ньютона–Лейбница.

а)

$$\int_{1}^{2}\left(\frac{3}{x^2}+x^2+2\right)\,dx = \int_{1}^{2}\left(3x^{-2}+x^2+2\right)\,dx$$
$$=\left(-\frac{3}{x}+\frac{x^3}{3}+2x\right)\Bigg|_{1}^{2}$$
$$=\left(-\frac{3}{2}+\frac{8}{3}+4\right)-\left(-3+\frac{1}{3}+2\right) =\frac{35}{6}-\left(-\frac{5}{3}\right) =\frac{45}{6} =5\frac{5}{6}.$$

б)

$$\int_{0}^{1}\left(\frac{2}{\sqrt{x+1}}-\frac{3}{(x+1)^2}\right)\,dx = \int_{0}^{1}\left(2(x+1)^{-\frac12}-3(x+1)^{-2}\right)\,dx$$
$$=\left(4\sqrt{x+1}+\frac{3}{x+1}\right)\Bigg|_{0}^{1}$$
$$=\left(4\sqrt2+\frac{3}{2}\right)-\left(4+\frac{3}{1}\right) =4\sqrt2-\frac{11}{2} =4\sqrt2-5{,}5.$$

в)

$$\int_{-2}^{-1}\left(-\frac{5}{x^2}+x^4-3x\right)\,dx = \int_{-2}^{-1}\left(-5x^{-2}+x^4-3x\right)\,dx$$
$$=\left(\frac{5}{x}+\frac{x^5}{5}-\frac{3x^2}{2}\right)\Bigg|_{-2}^{-1}$$
$$=\left(-5-\frac{1}{5}-\frac{3}{2}\right)-\left(-\frac{5}{2}-\frac{32}{5}-6\right) =8{,}2.$$

г)

$$\int_{5}^{8}\left(\frac{2}{(x-2)^2}-\frac{1}{\sqrt{x-4}}\right)\,dx = \int_{5}^{8}\left(2(x-2)^{-2}-(x-4)^{-\frac12}\right)\,dx$$
$$=\left(-\frac{2}{x-2}-2\sqrt{x-4}\right)\Bigg|_{5}^{8}$$
$$=\left(-\frac{2}{6}-2\sqrt4\right)-\left(-\frac{2}{3}-2\sqrt1\right) =-\frac{1}{3}-4+\frac{2}{3}+2 =-1\frac{1}{3}.$$

Ответ: а) $$5\frac{5}{6}$$; б) $$4\sqrt2-5{,}5$$; в) $$8{,}2$$; г) $$-1\frac{1}{3}$$.