11.4. Вычислите определённый интеграл: 1) $$int_{-4}^{-2} 2dx$$; 2) $$int_{-1}^{2} x^3 dx$$; 3) $$int_{0}^{\frac{\pi}{2}} cosx dx$$; 4) $$int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{sin^2 x}$$; 5) $$int_{1}^{3} \frac{dx}{x^4}$$; 6) $$int_{0}^{4} e^x dx$$; 7) $$int_{1}^{e} \frac{dx}{x}$$; 8) $$int_{4}^{9} \sqrt{x} dx$$; 9) $$int_{-1}^{1} (1-5x^4) dx$$;

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ФИО:Телефон:Емаил:Полное описание сути нарушения прав (почему распространение данной информации запрещено Правообладателем):

Accepted Answer

$$int_{-4}^{-2} 2dx = 2x \Big|_{-4}^{-2} = 2(-2) - 2(-4) = -4 + 8 = 4$$
$$int_{-1}^{2} x^3 dx = \frac{x^4}{4} \Big|_{-1}^{2} = \frac{2^4}{4} - \frac{(-1)^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} = 3.75$$
$$int_{0}^{\frac{\pi}{2}} cosx dx = sinx \Big|_{0}^{\frac{\pi}{2}} = sin(\frac{\pi}{2}) - sin(0) = 1 - 0 = 1$$
$$int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{sin^2 x} = -cotx \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -cot(\frac{\pi}{2}) - (-cot(\frac{\pi}{4})) = -0 + 1 = 1$$
$$int_{1}^{3} \frac{dx}{x^4} = \int_{1}^{3} x^{-4} dx = \frac{x^{-3}}{-3} \Big|_{1}^{3} = \frac{1}{-3x^3} \Big|_{1}^{3} = \frac{1}{-3(3^3)} - \frac{1}{-3(1^3)} = -\frac{1}{81} + \frac{1}{3} = \frac{-1 + 27}{81} = \frac{26}{81}$$
$$int_{0}^{4} e^x dx = e^x \Big|_{0}^{4} = e^4 - e^0 = e^4 - 1$$
$$int_{1}^{e} \frac{dx}{x} = ln|x| \Big|_{1}^{e} = ln|e| - ln|1| = 1 - 0 = 1$$
$$int_{4}^{9} \sqrt{x} dx = \int_{4}^{9} x^{\frac{1}{2}} dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \Big|_{4}^{9} = \frac{2}{3} x^{\frac{3}{2}} \Big|_{4}^{9} = \frac{2}{3} (9^{\frac{3}{2}}) - \frac{2}{3} (4^{\frac{3}{2}}) = \frac{2}{3} (27) - \frac{2}{3} (8) = 18 - \frac{16}{3} = \frac{54 - 16}{3} = \frac{38}{3}$$
$$int_{-1}^{1} (1-5x^4) dx = x - x^5 \Big|_{-1}^{1} = (1 - 1^5) - (-1 - (-1)^5) = (1 - 1) - (-1 - (-1)) = 0 - (-1 + 1) = 0 - 0 = 0$$