Упр.1539 ГДЗ Алимов 10-11 класс (Алгебра)

1. $$y=4x-x^2,\quad y=0,\quad x=0,\quad x=3$$
2. $$y=x^2-2x+8,\quad y=6,\quad x=-1,\quad x=3$$
3. $$y=\sin x,\quad y=0,\quad x=\frac{2\pi}{3},\quad x=\pi$$
4. $$y=\cos x,\quad y=0,\quad x=-\frac{\pi}{6},\quad x=\frac{\pi}{6}$$

1) $$S=\int_0^3\bigl((4x-x^2)-5\bigr)\,dx=\int_0^3(4x-x^2-5)\,dx$$

$$S=\left(2x^2-\frac{x^3}{3}-5x\right)\Bigg|_0^3 = \left(18-9-15\right)-0=-6$$

Площадь равна модулю полученного значения:

$$S=6.$$

2) $$S=\int_{-1}^3\bigl((x^2-2x+8)-6\bigr)\,dx=\int_{-1}^3(x^2-2x+2)\,dx$$

$$S=\left(\frac{x^3}{3}-x^2+2x\right)\Bigg|_{-1}^3 =\left(9-9+6\right)-\left(-\frac13-1-2\right) =6+\frac{10}{3}=\frac{28}{3}$$

$$S=9\frac13.$$

3) $$S=\int_{2\pi/3}^{\pi}\sin x\,dx=-\cos x\Bigg|_{2\pi/3}^{\pi}$$

$$S=-\cos\pi+\cos\frac{2\pi}{3} =1-\frac12=\frac12$$

4) $$S=\int_{-\pi/6}^{\pi/6}\cos x\,dx=\sin x\Bigg|_{-\pi/6}^{\pi/6}$$

$$S=\sin\frac{\pi}{6}-\sin\left(-\frac{\pi}{6}\right) =\frac12+\frac12=1$$

Ответ

1) $$6$$; 2) $$9\frac13$$; 3) $$\frac12$$; 4) $$1$$.