Упр.363 ГДЗ Колмогоров 10-11 класс (Алгебра)

10–11

а)

$$\left(\sin \frac{x}{4}+\cos \frac{x}{4}\right)^2 = \sin^2 \frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}+\cos^2 \frac{x}{4} = 1+\sin \frac{x}{2}.$$

Тогда
$$\int_0^{\frac{2\pi}{3}}\left(\sin \frac{x}{4}+\cos \frac{x}{4}\right)^2\,dx = \int_0^{\frac{2\pi}{3}}\left(1+\sin \frac{x}{2}\right)\,dx = \left(x-2\cos \frac{x}{2}\right)\Bigg|_0^{\frac{2\pi}{3}}.$$

$$\left(\frac{2\pi}{3}-2\cos \frac{\pi}{3}\right)-\left(0-2\cos 0\right) = \frac{2\pi}{3}-1+2 = \frac{2\pi}{3}+1.$$

б)

$$\int_0^2 (1+2x)^3\,dx = \frac{(1+2x)^4}{8}\Bigg|_0^2 = \frac{5^4-1^4}{8} = \frac{625-1}{8} = 78.$$

в)

$$\int_0^{\frac{\pi}{12}} (1+\cos 2x)\,dx = \left(x+\frac12 \sin 2x\right)\Bigg|_0^{\frac{\pi}{12}} = \frac{\pi}{12}+\frac12 \sin \frac{\pi}{6}.$$

$$\frac{\pi}{12}+\frac12 \cdot \frac12 = \frac{\pi}{12}+\frac14.$$

г)

$$\int_1^4 \left(x+\frac{\sqrt{x}}{x}\right)\,dx = \int_1^4 \left(x+x^{-1/2}\right)\,dx = \left(\frac{x^2}{2}+2\sqrt{x}\right)\Bigg|_1^4.$$

$$\left(\frac{16}{2}+2\sqrt{4}\right)-\left(\frac{1}{2}+2\sqrt{1}\right) = 8+4-\frac12-2 = \frac{19}{2}.$$

Ответ

$$\text{а) } \frac{2\pi}{3}+1;\quad \text{б) } 78;\quad \text{в) } \frac{\pi}{12}+\frac14;\quad \text{г) } \frac{19}{2}.$$