Упр.433 Повторение ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)
1) y=x^3+1, y=0, x=0, x=2; 6) y=4/x^2, y=x-1, x=1;
2) y=2-x^2, y=0; 7) y=x^2-4x+5, y=5-x;
3) y=1/x, y=0, x=1, x=3; 8) y=8-x^2, y=4;
4) y=e^(-x), y=1, x=-2; 9) y=x^2, y=4x-x^2;
5) y=-x^2+4, x+y=4; 10) y=5/x, y=4x+1, x=2.
$$S=\int_0^2 (x^3+1)\,dx=\left(\frac{x^4}{4}+x\right)\Bigg|_0^2$$
$$S=\left(\frac{16}{4}+2\right)-\left(\frac{0}{4}+0\right)=4+2=6.$$
Точки пересечения с осью $$Ox$$:
$$2-x^2=0,\quad x=\pm\sqrt2.$$
$$S=\int_{-\sqrt2}^{\sqrt2}(2-x^2)\,dx=\left(2x-\frac{x^3}{3}\right)\Bigg|_{-\sqrt2}^{\sqrt2}$$
$$S=\left(2\sqrt2-\frac{2\sqrt2}{3}\right)-\left(-2\sqrt2+\frac{2\sqrt2}{3}\right)=\frac{8\sqrt2}{3}.$$
$$S=\int_1^3 \frac{dx}{x}=\ln|x|\Bigg|_1^3=\ln 3.$$
Точка пересечения:
$$e^{-x}=1,\quad x=0.$$
$$S=\int_{-2}^0 (e^{-x}-1)\,dx=\left(-e^{-x}-x\right)\Bigg|_{-2}^0$$
$$S=(-1-0)-(-e^2+2)=e^2-3.$$
Точки пересечения:
$$-x^2+4=4-x,\quad x(x-1)=0,\quad x_1=0,\ x_2=1.$$
$$S=\int_0^1 \bigl((-x^2+4)-(4-x)\bigr)\,dx=\int_0^1 (x-x^2)\,dx$$
$$S=\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\Bigg|_0^1=\frac12-\frac13=\frac16.$$
Точка пересечения:
$$\frac{4}{x^2}=x-1,\quad 4=x^3-x^2,\quad (x-2)(x^2+x+2)=0,\quad x=2.$$
$$S=\int_1^2 \left(\frac{4}{x^2}-(x-1)\right)\,dx=\int_1^2 \left(\frac{4}{x^2}-x+1\right)\,dx$$
$$S=\left(-\frac{4}{x}-\frac{x^2}{2}+x\right)\Bigg|_1^2$$
$$S=\left(-2-2+2\right)-\left(-4-\frac12+1\right)=\frac32.$$
Точки пересечения:
$$x^2-4x+5=5-x,\quad x(x-3)=0,\quad x_1=0,\ x_2=3.$$
$$S=\int_0^3 \bigl((x^2-4x+5)-(5-x)\bigr)\,dx=\int_0^3 (x^2-3x)\,dx$$
$$S=\left(\frac{x^3}{3}-\frac{3x^2}{2}\right)\Bigg|_0^3=\frac{27}{3}-\frac{27}{2}=\frac{9}{2}.$$
Точки пересечения:
$$8-x^2=4,\quad x=\pm2.$$
$$S=\int_{-2}^2 \bigl((8-x^2)-4\bigr)\,dx=\int_{-2}^2 (4-x^2)\,dx$$
$$S=\left(4x-\frac{x^3}{3}\right)\Bigg|_{-2}^2=\frac{32}{3}.$$
Точки пересечения:
$$x^2=4x-x^2,\quad 2x(x-2)=0,\quad x_1=0,\ x_2=2.$$
$$S=\int_0^2 \bigl((4x-x^2)-x^2\bigr)\,dx=\int_0^2 (4x-2x^2)\,dx$$
$$S=\left(2x^2-\frac{2x^3}{3}\right)\Bigg|_0^2=\frac{16}{3}.$$
Точки пересечения:
$$\frac{5}{x}=4x+1,\quad 4x^2+x-5=0,$$
$$D=1+80=81,\quad x_1=-\frac54,\ x_2=1.$$
С учётом границы $$x=2$$:
$$S=\int_1^2 \left(4x+1-\frac{5}{x}\right)\,dx=\left(2x^2+x-5\ln|x|\right)\Bigg|_1^2$$
$$S=(8+2-5\ln2)-(2+1-5\ln1)=7-5\ln2.$$
Ответ
1) $$6$$; 2) $$\frac{8\sqrt2}{3}$$; 3) $$\ln 3$$; 4) $$e^2-3$$; 5) $$\frac16$$; 6) $$\frac32$$; 7) $$\frac92$$; 8) $$\frac{32}{3}$$; 9) $$\frac{16}{3}$$; 10) $$7-5\ln2$$.
