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

Задача

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

Подробный ответ

Вычислим значения определённых интегралов.

а)

$$\int_{\pi}^{\frac{3\pi}{2}} \cos\left(1{,}5\pi+0{,}5x\right)\,dx =2\sin\left(1{,}5\pi+0{,}5x\right)\Bigg|_{\pi}^{\frac{3\pi}{2}}$$
$$=2\sin\left(\frac{3\pi}{2}+\frac{3\pi}{4}\right)-2\sin\left(\frac{3\pi}{2}+\frac{\pi}{2}\right) =2\sin\frac{9\pi}{4}-2\sin 2\pi$$
$$=2\sin\left(2\pi+\frac{\pi}{4}\right)-0 =2\cdot\frac{\sqrt{2}}{2} =\sqrt{2}.$$

б)

$$\int_{1}^{2}\left(x^{-2}+x^2\right)\,dx =\left(-x^{-1}+\frac{x^3}{3}\right)\Bigg|_{1}^{2} =\left(\frac{8}{3}-\frac12\right)-\left(\frac13-1\right)$$
$$=\frac{16-3}{6}-\left(-\frac23\right) =\frac{13}{6}+\frac{4}{6} =\frac{17}{6}.$$

в)

$$\int_{\frac{\pi}{12}}^{\frac{\pi}{6}}(\cos 3x-\sin 2x)\,dx =\left(\frac13\sin 3x+\frac12\cos 2x\right)\Bigg|_{\frac{\pi}{12}}^{\frac{\pi}{6}}$$
$$=\left(\frac13\sin\frac{\pi}{2}+\frac12\cos\frac{\pi}{3}\right) -\left(\frac13\sin\frac{\pi}{4}+\frac12\cos\frac{\pi}{6}\right)$$
$$=\left(\frac13+\frac14\right)-\left(\frac{\sqrt2}{6}+\frac{\sqrt3}{4}\right) =\frac{7}{12}-\frac{2\sqrt2}{12}-\frac{3\sqrt3}{12} =\frac{7-2\sqrt2-3\sqrt3}{12}.$$

г)

$$\int_{-5}^{-2}(5-6x-x^2)\,dx =\left(5x-3x^2-\frac{x^3}{3}\right)\Bigg|_{-5}^{-2}$$
$$=\left(5\cdot(-2)-3\cdot4-\frac{(-2)^3}{3}\right) -\left(5\cdot(-5)-3\cdot25-\frac{(-5)^3}{3}\right)$$
$$=\left(-10-12+\frac{8}{3}\right)-\left(-25-75+\frac{125}{3}\right) =-\frac{58}{3}-\left(-\frac{175}{3}\right) =\frac{117}{3}=39.$$