11. Вычислите интегралы: 1) $$\int_{-1}^{2} dx$$; 2) $$\int_{0}^{3} 5dx$$; 3) $$\int_{-2}^{3} xdx$$; 4) $$\int_{0}^{1} x^2 dx$$; 5) $$\int_{1}^{3} (3-2x)dx$$; 6) $$\int_{0}^{1} (x^2+1)dx$$; 7) $$\int_{-1}^{0} (x^2+2x)dx$$; 8) $$\int_{-1}^{1} (2x^3-x-1)dx$$; 9) $$\int_{-1}^{2} (2x^2-5x-7)dx$$; 10) $$\int_{1}^{2} \frac{dx}{x^2}$$; 11) $$\int_{1}^{2} \frac{dx}{3x^2}$$; 12) $$\int_{1}^{4} \sqrt{x} dx$$; 13) $$\int_{0}^{1} \sqrt[3]{x} dx$$; 14) $$\int_{2}^{8} \frac{dx}{\sqrt{x}}$$; 15) $$\int_{1}^{2} \frac{2x^2+3x-2}{x^3} dx$$; 16) $$\int_{1}^{8} \frac{2x^2+\sqrt{x}+1}{\sqrt{x}} dx$$; 17) $$\int_{0}^{\frac{\pi}{2}} \sin 5x dx$$; 18) $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x dx$$; 19) $$\int_{0}^{\frac{\pi}{4}} \sin \frac{x}{2}\cos \frac{x}{2} dx$$; 20) $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \frac{x}{2} dx$$; 21) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^2 x}$$; 22) $$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^2 x}$$; 23) $$\int_{0}^{\pi} (\sin x-3\cos x-x) dx$$; 24) $$\int_{0}^{\frac{\pi}{4}} tg^2 x dx$$;

Question

11. Вычислите интегралы: 1) $$\int_{-1}^{2} dx$$; 2) $$\int_{0}^{3} 5dx$$; 3) $$\int_{-2}^{3} xdx$$; 4) $$\int_{0}^{1} x^2 dx$$; 5) $$\int_{1}^{3} (3-2x)dx$$; 6) $$\int_{0}^{1} (x^2+1)dx$$; 7) $$\int_{-1}^{0} (x^2+2x)dx$$; 8) $$\int_{-1}^{1} (2x^3-x-1)dx$$; 9) $$\int_{-1}^{2} (2x^2-5x-7)dx$$; 10) $$\int_{1}^{2} \frac{dx}{x^2}$$; 11) $$\int_{1}^{2} \frac{dx}{3x^2}$$; 12) $$\int_{1}^{4} \sqrt{x} dx$$; 13) $$\int_{0}^{1} \sqrt[3]{x} dx$$; 14) $$\int_{2}^{8} \frac{dx}{\sqrt{x}}$$; 15) $$\int_{1}^{2} \frac{2x^2+3x-2}{x^3} dx$$; 16) $$\int_{1}^{8} \frac{2x^2+\sqrt{x}+1}{\sqrt{x}} dx$$; 17) $$\int_{0}^{\frac{\pi}{2}} \sin 5x dx$$; 18) $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x dx$$; 19) $$\int_{0}^{\frac{\pi}{4}} \sin \frac{x}{2}\cos \frac{x}{2} dx$$; 20) $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \frac{x}{2} dx$$; 21) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^2 x}$$; 22) $$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^2 x}$$; 23) $$\int_{0}^{\pi} (\sin x-3\cos x-x) dx$$; 24) $$\int_{0}^{\frac{\pi}{4}} tg^2 x dx$$;

$Фото задания: 11. Вычислите интегралы: 1) $\int_{-1}^{2} dx$; 2) $\int_{0}^{3} 5dx$; 3) $\int_{-2}^{3} xdx$; 4) $\int_{0}^{1} x^2 dx$; 5)…$ Фото задания

Accepted Answer

Решение:

$$\int_{-1}^{2} dx = [x]_{-1}^{2} = 2 - (-1) = 3$$
$$\int_{0}^{3} 5dx = [5x]_{0}^{3} = 5 \cdot 3 - 5 \cdot 0 = 15$$
$$\int_{-2}^{3} xdx = [\frac{x^2}{2}]_{-2}^{3} = \frac{3^2}{2} - \frac{(-2)^2}{2} = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}$$
$$\int_{0}^{1} x^2 dx = [\frac{x^3}{3}]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$$
$$\int_{1}^{3} (3-2x)dx = [3x - x^2]_{1}^{3} = (3 \cdot 3 - 3^2) - (3 \cdot 1 - 1^2) = (9 - 9) - (3 - 1) = 0 - 2 = -2$$
$$\int_{0}^{1} (x^2+1)dx = [\frac{x^3}{3} + x]_{0}^{1} = (\frac{1^3}{3} + 1) - (\frac{0^3}{3} + 0) = \frac{1}{3} + 1 = \frac{4}{3}$$
$$\int_{-1}^{0} (x^2+2x)dx = [\frac{x^3}{3} + x^2]_{-1}^{0} = (\frac{0^3}{3} + 0^2) - (\frac{(-1)^3}{3} + (-1)^2) = 0 - (-\frac{1}{3} + 1) = -\frac{2}{3}$$
$$\int_{-1}^{1} (2x^3-x-1)dx = [\frac{2x^4}{4} - \frac{x^2}{2} - x]_{-1}^{1} = [\frac{x^4}{2} - \frac{x^2}{2} - x]_{-1}^{1} = (\frac{1^4}{2} - \frac{1^2}{2} - 1) - (\frac{(-1)^4}{2} - \frac{(-1)^2}{2} - (-1)) = (\frac{1}{2} - \frac{1}{2} - 1) - (\frac{1}{2} - \frac{1}{2} + 1) = -1 - 1 = -2$$
$$\int_{-1}^{2} (2x^2-5x-7)dx = [\frac{2x^3}{3} - \frac{5x^2}{2} - 7x]_{-1}^{2} = (\frac{2(2)^3}{3} - \frac{5(2)^2}{2} - 7(2)) - (\frac{2(-1)^3}{3} - \frac{5(-1)^2}{2} - 7(-1)) = (\frac{16}{3} - 10 - 14) - (-\frac{2}{3} - \frac{5}{2} + 7) = (\frac{16}{3} - 24) - (-\frac{2}{3} - \frac{5}{2} + 7) = \frac{16-72}{3} - \frac{-4-15+42}{6} = \frac{-56}{3} - \frac{23}{6} = \frac{-112-23}{6} = \frac{-135}{6} = -\frac{45}{2}$$
$$\int_{1}^{2} \frac{dx}{x^2} = [-\frac{1}{x}]_{1}^{2} = -\frac{1}{2} - (-\frac{1}{1}) = -\frac{1}{2} + 1 = \frac{1}{2}$$
$$\int_{1}^{2} \frac{dx}{3x^2} = [-\frac{1}{3x}]_{1}^{2} = -\frac{1}{3 \cdot 2} - (-\frac{1}{3 \cdot 1}) = -\frac{1}{6} + \frac{1}{3} = \frac{-1+2}{6} = \frac{1}{6}$$
$$\int_{1}^{4} \sqrt{x} dx = \int_{1}^{4} x^{\frac{1}{2}} dx = [\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]_{1}^{4} = [\frac{2}{3} x^{\frac{3}{2}}]_{1}^{4} = \frac{2}{3} (4^{\frac{3}{2}} - 1^{\frac{3}{2}}) = \frac{2}{3} (8 - 1) = \frac{2}{3} \cdot 7 = \frac{14}{3}$$
$$\int_{0}^{1} \sqrt[3]{x} dx = \int_{0}^{1} x^{\frac{1}{3}} dx = [\frac{x^{\frac{4}{3}}}{\frac{4}{3}}]_{0}^{1} = [\frac{3}{4} x^{\frac{4}{3}}]_{0}^{1} = \frac{3}{4} (1^{\frac{4}{3}} - 0^{\frac{4}{3}}) = \frac{3}{4} \cdot 1 = \frac{3}{4}$$
$$\int_{2}^{8} \frac{dx}{\sqrt{x}} = \int_{2}^{8} x^{-\frac{1}{2}} dx = [\frac{x^{\frac{1}{2}}}{\frac{1}{2}}]_{2}^{8} = [2 \sqrt{x}]_{2}^{8} = 2 \sqrt{8} - 2 \sqrt{2} = 2(2\sqrt{2}) - 2\sqrt{2} = 4\sqrt{2} - 2\sqrt{2} = 2\sqrt{2}$$
$$\int_{1}^{2} \frac{2x^2+3x-2}{x^3} dx = \int_{1}^{2} (\frac{2}{x} + \frac{3}{x^2} - \frac{2}{x^3}) dx = [2 \ln|x| - \frac{3}{x} - \frac{2x^{-2}}{-2}]_{1}^{2} = [2 \ln|x| - \frac{3}{x} + \frac{1}{x^2}]_{1}^{2} = (2 \ln 2 - \frac{3}{2} + \frac{1}{4}) - (2 \ln 1 - \frac{3}{1} + \frac{1}{1^2}) = (2 \ln 2 - \frac{6}{4} + \frac{1}{4}) - (0 - 3 + 1) = 2 \ln 2 - \frac{5}{4} - (-2) = 2 \ln 2 - \frac{5}{4} + 2 = 2 \ln 2 + \frac{3}{4}$$
$$\int_{1}^{8} \frac{2x^2+\sqrt{x}+1}{\sqrt{x}} dx = \int_{1}^{8} (2x^{\frac{3}{2}} + 1 + x^{-\frac{1}{2}}) dx = [2\frac{x^{\frac{5}{2}}}{\frac{5}{2}} + x + \frac{x^{\frac{1}{2}}}{\frac{1}{2}}]_{1}^{8} = [\frac{4}{5}x^{\frac{5}{2}} + x + 2\sqrt{x}]_{1}^{8} = (\frac{4}{5}(8^{\frac{5}{2}}) + 8 + 2\sqrt{8}) - (\frac{4}{5}(1^{\frac{5}{2}}) + 1 + 2\sqrt{1}) = (\frac{4}{5}(128\sqrt{2}) + 8 + 4\sqrt{2}) - (\frac{4}{5} + 1 + 2) = \frac{512\sqrt{2}}{5} + 8 + 4\sqrt{2} - \frac{4}{5} - 3 = \frac{512\sqrt{2}}{5} + 5 + 4\sqrt{2} - \frac{4}{5} = \frac{512\sqrt{2} + 25 + 20\sqrt{2} - 4}{5} = \frac{532\sqrt{2} + 21}{5}$$
$$\int_{0}^{\frac{\pi}{2}} \sin 5x dx = [-\frac{\cos 5x}{5}]_{0}^{\frac{\pi}{2}} = -\frac{\cos (5\frac{\pi}{2})}{5} - (-\frac{\cos 0}{5}) = -\frac{0}{5} - (-\frac{1}{5}) = \frac{1}{5}$$
$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x dx = [\sin x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \sin \frac{\pi}{3} - \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$$
$$\int_{0}^{\frac{\pi}{4}} \sin \frac{x}{2}\cos \frac{x}{2} dx = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sin x dx = \frac{1}{2} [-\cos x]_{0}^{\frac{\pi}{4}} = -\frac{1}{2} (\cos \frac{\pi}{4} - \cos 0) = -\frac{1}{2} (\frac{\sqrt{2}}{2} - 1) = \frac{1}{2} (1 - \frac{\sqrt{2}}{2}) = \frac{2-\sqrt{2}}{4}$$
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \frac{x}{2} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos x}{2} dx = [\frac{1}{2}x + \frac{1}{2}\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = (\frac{1}{2}\frac{\pi}{2} + \frac{1}{2}\sin \frac{\pi}{2}) - (\frac{1}{2}(-\frac{\pi}{2}) + \frac{1}{2}\sin (-\frac{\pi}{2})) = (\frac{\pi}{4} + \frac{1}{2}) - (-\frac{\pi}{4} - \frac{1}{2}) = \frac{\pi}{4} + \frac{1}{2} + \frac{\pi}{4} + \frac{1}{2} = \frac{\pi}{2} + 1$$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^2 x} = [-\text{ctg } x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\text{ctg } \frac{\pi}{2} - (-\text{ctg } \frac{\pi}{4}) = -(0) - (-1) = 1$$
$$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^2 x} = [\text{tg } x]_{0}^{\frac{\pi}{4}} = \text{tg } \frac{\pi}{4} - \text{tg } 0 = 1 - 0 = 1$$
$$\int_{0}^{\pi} (\sin x-3\cos x-x) dx = [-\cos x - 3\sin x - \frac{x^2}{2}]_{0}^{\pi} = (-\cos \pi - 3\sin \pi - \frac{\pi^2}{2}) - (-\cos 0 - 3\sin 0 - \frac{0^2}{2}) = (-(-1) - 3(0) - \frac{\pi^2}{2}) - (-1 - 0 - 0) = (1 - \frac{\pi^2}{2}) - (-1) = 1 - \frac{\pi^2}{2} + 1 = 2 - \frac{\pi^2}{2}$$
$$\int_{0}^{\frac{\pi}{4}} tg^2 x dx = \int_{0}^{\frac{\pi}{4}} (\sec^2 x - 1) dx = [\text{tg } x - x]_{0}^{\frac{\pi}{4}} = (\text{tg } \frac{\pi}{4} - \frac{\pi}{4}) - (\text{tg } 0 - 0) = (1 - \frac{\pi}{4}) - 0 = 1 - \frac{\pi}{4}$$

Ответ: 1) 3; 2) 15; 3) $$\frac{5}{2}$$; 4) $$\frac{1}{3}$$; 5) -2; 6) $$\frac{4}{3}$$; 7) -$$\frac{2}{3}$$; 8) -2; 9) -$$\frac{45}{2}$$; 10) $$\frac{1}{2}$$; 11) $$\frac{1}{6}$$; 12) $$\frac{14}{3}$$; 13) $$\frac{3}{4}$$; 14) $$2\sqrt{2}$$; 15) $$2 \ln 2 + \frac{3}{4}$$; 16) $$\frac{532\sqrt{2} + 21}{5}$$; 17) $$\frac{1}{5}$$; 18) $$\frac{\sqrt{3}-1}{2}$$; 19) $$\frac{2-\sqrt{2}}{4}$$; 20) $$\frac{\pi}{2} + 1$$; 21) 1; 22) 1; 23) $$2 - \frac{\pi^2}{2}$$; 24) $$1 - \frac{\pi}{4}$$.