Упр.364 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) $$S=\int\limits_{1}^{2} (8-x^3)\,dx$$

$$S=\left(8x-\frac{x^4}{4}\right)\Bigg|_{1}^{2}= \left(16-\frac{16}{4}\right)-\left(8-\frac14\right)=8-\frac{31}{4}=\frac{1}{4}$$

$$S=4\frac14$$

б) $$S=\int\limits_{-\pi/3}^{\pi/3} (2\cos x-1)\,dx$$

$$S=\left(2\sin x-x\right)\Bigg|_{-\pi/3}^{\pi/3}=2\sin\frac{\pi}{3}-\frac{\pi}{3}-2\sin\left(-\frac{\pi}{3}\right)+\frac{\pi}{3}$$

$$S=2\cdot\frac{\sqrt3}{2}+2\cdot\frac{\sqrt3}{2}-\frac{2\pi}{3}=2\sqrt3-\frac{2\pi}{3}$$

в) Найдём точку пересечения параболы и прямой:

$$x^2-2x+4=3$$

$$x^2-2x+1=0$$

$$x=1$$

Тогда площадь равна

$$S=\int\limits_{-1}^{1} (x^2-2x+4)\,dx-2\cdot 3$$

$$S=\left(\frac{x^3}{3}-x^2+4x\right)\Bigg|_{-1}^{1}-6$$

$$S=\left(\frac13-1+4\right)-\left(-\frac13-1-4\right)-6=\frac{8}{3}$$

г) $$S=\int\limits_{\pi/6}^{5\pi/6}\left(\sin x-\frac12\right)\,dx$$

$$S=\left(-\cos x-\frac{x}{2}\right)\Bigg|_{\pi/6}^{5\pi/6}$$

$$S=-\cos\frac{5\pi}{6}-\frac{5\pi}{12}+\cos\frac{\pi}{6}+\frac{\pi}{12}$$

$$S=\frac{\sqrt3}{2}-\left(-\frac{\sqrt3}{2}\right)-\frac{4\pi}{12}=\sqrt3-\frac{\pi}{3}$$

Ответ

а) $$4\frac14$$; б) $$2\sqrt3-\frac{2\pi}{3}$$; в) $$\frac{8}{3}$$; г) $$\sqrt3-\frac{\pi}{3}$$.