Упр.432 Повторение ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)

Задача

Вычислите интеграл:

$$\int\limits_{1}^{3}\frac{dx}{x^2}$$;
$$\int\limits_{0}^{\pi}(6\cos 4x-3\sin x)\,dx$$;
$$\int\limits_{0}^{\frac{\pi}{18}}\frac{dx}{\left(\sin\left(3x+\frac{\pi}{6}\right)\right)^2}$$;
$$\int\limits_{-2}^{1}(x^2-2x+4)\,dx$$;
$$\int\limits_{1}^{3}\left(\frac{4}{x}-x\right)\,dx$$;
$$\int\limits_{-2}^{2}\frac{dx}{\sqrt{2x+5}}$$;
$$\int\limits_{0}^{2}(3x-2)^3\,dx$$;
$$\int\limits_{2}^{4}e^{-x}\,dx$$;
$$\int\limits_{0}^{5}\frac{dx}{4x+1}$$.

Подробный ответ

$$\int_1^3 \frac{dx}{x^2}=\int_1^3 x^{-2}\,dx=-\frac{1}{x}\Big|_1^3=-\frac13-(-1)=\frac23.$$
$$\int_0^\pi (6\cos 4x-3\sin x)\,dx=\left(\frac{6}{4}\sin 4x+3\cos x\right)\Big|_0^\pi.$$
$$\left(\frac32\sin 4\pi+3\cos\pi\right)-\left(\frac32\sin 0+3\cos 0\right)=-3-3=-6.$$
$$\int_0^{\pi/18}\frac{dx}{\sin^2\left(3x+\frac{\pi}{6}\right)}=-\frac13\ctg\left(3x+\frac{\pi}{6}\right)\Big|_0^{\pi/18}.$$
$$-\frac13\ctg\frac{\pi}{3}+\frac13\ctg\frac{\pi}{6}=-\frac13\cdot\frac{\sqrt3}{3}+\frac13\cdot\sqrt3=\frac{2\sqrt3}{9}.$$
$$\int_{-2}^1 (x^2-2x+4)\,dx=\left(\frac{x^3}{3}-x^2+4x\right)\Big|_{-2}^1.$$
$$\left(\frac13-1+4\right)-\left(-\frac83-4-8\right)=\frac{10}{3}+\frac{44}{3}=18.$$
$$\int_1^3 \left(\frac{4}{x}-x\right)\,dx=\left(4\ln|x|-\frac{x^2}{2}\right)\Big|_1^3.$$
$$\left(4\ln 3-\frac92\right)-\left(4\ln 1-\frac12\right)=4\ln 3-4.$$
$$\int_{-2}^2 \frac{dx}{\sqrt{2x+5}}.$$
Сделаем замену $$u=2x+5,$$ тогда $$du=2dx,\quad dx=\frac{du}{2}.$$
$$\int \frac{dx}{\sqrt{2x+5}}=\sqrt{2x+5}.$$
Значит,
$$\int_{-2}^2 \frac{dx}{\sqrt{2x+5}}=\sqrt{9}-\sqrt{1}=2.$$
$$\int_0^2 (3x-2)^3\,dx=\frac{(3x-2)^4}{12}\Big|_0^2.$$
$$\frac{(6-2)^4}{12}-\frac{(0-2)^4}{12}=\frac{256-16}{12}=20.$$
$$\int_2^4 e^{-x}\,dx=-e^{-x}\Big|_2^4=e^{-2}-e^{-4}=\frac{e^2-1}{e^4}.$$
$$\int_0^5 \frac{dx}{4x+1}=\frac14\ln|4x+1|\Big|_0^5.$$
$$\frac14\ln 21-\frac14\ln 1=\frac14\ln 21.$$