Упр.273 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

1. Вычислите: а) $$\int_{\pi}^{\frac{3\pi}{2}}\cos(1{,}5\pi+0{,}5x)\,dx$$; б) $$\int_{1}^{2}(x^{-2}+x^2)\,dx$$; в) $$\int_{\frac{\pi}{12}}^{\frac{\pi}{6}}\cos(3x-\sin(2x))\,dx$$; г) $$\int_{-5}^{-2}(5-6x-x^2)\,dx$$.

а) $$\int\limits_{\pi}^{\frac{3\pi}{2}} \cos\left(1{,}5\pi+0{,}5x\right)\,dx$$

Сделаем замену: $$u=1{,}5\pi+0{,}5x,\quad du=0{,}5\,dx,\quad dx=2\,du.$$

Тогда

$$\int\limits_{\pi}^{\frac{3\pi}{2}} \cos\left(1{,}5\pi+0{,}5x\right)\,dx=2\sin\left(1{,}5\pi+0{,}5x\right)\Bigg|_{\pi}^{\frac{3\pi}{2}}.$$

Подставим пределы:

$$2\sin\left(\frac{3\pi}{2}+\frac{3\pi}{4}\right)-2\sin\left(\frac{3\pi}{2}+\frac{\pi}{2}\right)=2\sin\frac{9\pi}{4}-2\sin 2\pi=\sqrt{2}.$$

б) $$\int\limits_{1}^{2}\left(x^{-2}+x^2\right)\,dx$$

$$\int\limits_{1}^{2}\left(x^{-2}+x^2\right)\,dx=\left(-x^{-1}+\frac{x^3}{3}\right)\Bigg|_{1}^{2}.$$

$$\left(-\frac12+\frac{8}{3}\right)-\left(-1+\frac13\right)=\frac{13}{6}+\frac{2}{3}=\frac{17}{6}.$$

в) $$\int\limits_{\frac{\pi}{12}}^{\frac{\pi}{6}} \left(\cos 3x-\sin 2x\right)\,dx$$

Первообразная:

$$\int \left(\cos 3x-\sin 2x\right)\,dx=\frac13\sin 3x+\frac12\cos 2x.$$

Тогда

$$\left(\frac13\sin 3x+\frac12\cos 2x\right)\Bigg|_{\frac{\pi}{12}}^{\frac{\pi}{6}}$$

$$=\left(\frac13\sin\frac{\pi}{2}+\frac12\cos\frac{\pi}{3}\right)-\left(\frac13\sin\frac{\pi}{4}+\frac12\cos\frac{\pi}{6}\right)$$

$$=\left(\frac13+\frac14\right)-\left(\frac{\sqrt2}{6}+\frac{\sqrt3}{4}\right)=\frac{7-2\sqrt2-3\sqrt3}{12}.$$

г) $$\int\limits_{-5}^{-2}\left(5-6x-x^2\right)\,dx$$

$$\int \left(5-6x-x^2\right)\,dx=5x-3x^2-\frac{x^3}{3}.$$

Тогда

$$\left(5x-3x^2-\frac{x^3}{3}\right)\Bigg|_{-5}^{-2}$$

$$=\left(-10-12+\frac{8}{3}\right)-\left(-25-75+\frac{125}{3}\right)=39.$$

Ответ

а) $$\sqrt2$$; б) $$\frac{17}{6}$$; в) $$\frac{7-2\sqrt2-3\sqrt3}{12}$$; г) $$39$$.