Вопрос:

Вычисли определённые интегралы: 1) $\int_{5}^{7} xdx$

Фотография

Ответ ассистента

Решим несколько определённых интегралов из твоего списка: 1) $\int_{5}^{7} x dx = \frac{x^2}{2} \Big|_{5}^{7} = \frac{7^2}{2} - \frac{5^2}{2} = \frac{49}{2} - \frac{25}{2} = \frac{24}{2} = 12$ 2) $\int_{3}^{8} dx = x \Big|_{3}^{8} = 8 - 3 = 5$ 3) $\int_{-3}^{0} x^2 dx = \frac{x^3}{3} \Big|_{-3}^{0} = \frac{0^3}{3} - \frac{(-3)^3}{3} = 0 - \frac{-27}{3} = 9$ 4) $\int_{-1}^{2} x^4 dx = \frac{x^5}{5} \Big|_{-1}^{2} = \frac{2^5}{5} - \frac{(-1)^5}{5} = \frac{32}{5} - \frac{-1}{5} = \frac{33}{5} = 6.6$ 5) $\int_{0}^{\frac{\pi}{2}} \sin(x) dx = -\cos(x) \Big|_{0}^{\frac{\pi}{2}} = -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 - (-1) = 1$ 6) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{dx}{\cos^2(x)} = \tan(x) \Big|_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \tan(\frac{\pi}{3}) - \tan(\frac{\pi}{4}) = \sqrt{3} - 1 \approx 0.732$ 7) $\int_{16}^{100} \frac{dx}{\sqrt{x}} = 2\sqrt{x} \Big|_{16}^{100} = 2\sqrt{100} - 2\sqrt{16} = 2 \cdot 10 - 2 \cdot 4 = 20 - 8 = 12$ 8) $\int_{e}^{e^3} \frac{dx}{x} = \ln|x| \Big|_{e}^{e^3} = \ln(e^3) - \ln(e) = 3 - 1 = 2$ 9) $\int_{1}^{10} \frac{dx}{x^2} = -\frac{1}{x} \Big|_{1}^{10} = -\frac{1}{10} - (-\frac{1}{1}) = -0.1 + 1 = 0.9$ 10) $\int_{-2}^{3} 3^x dx = \frac{3^x}{\ln(3)} \Big|_{-2}^{3} = \frac{3^3}{\ln(3)} - \frac{3^{-2}}{\ln(3)} = \frac{27}{\ln(3)} - \frac{1/9}{\ln(3)} = \frac{27 - 1/9}{\ln(3)} = \frac{242}{9\ln(3)} \approx 24.50$ 11) $\int_{1}^{8} \sqrt[3]{x} dx = \int_{1}^{8} x^{\frac{1}{3}} dx = \frac{3}{4}x^{\frac{4}{3}} \Big|_{1}^{8} = \frac{3}{4}(8^{\frac{4}{3}}) - \frac{3}{4}(1^{\frac{4}{3}}) = \frac{3}{4}(16) - \frac{3}{4}(1) = 12 - \frac{3}{4} = 11.25$ 12) $\int_{-4}^{-2} (2x + 4) dx = x^2 + 4x \Big|_{-4}^{-2} = ((-2)^2 + 4(-2)) - ((-4)^2 + 4(-4)) = (4 - 8) - (16 - 16) = -4 - 0 = -4$ 13) $\int_{0}^{6} (3x^2 - x) dx = x^3 - \frac{x^2}{2} \Big|_{0}^{6} = (6^3 - \frac{6^2}{2}) - (0^3 - \frac{0^2}{2}) = (216 - 18) - 0 = 198$ 14) $\int_{0}^{\frac{\pi}{2}} (4\sin(x) + 2\cos(x)) dx = -4\cos(x) + 2\sin(x) \Big|_{0}^{\frac{\pi}{2}} = (-4\cos(\frac{\pi}{2}) + 2\sin(\frac{\pi}{2})) - (-4\cos(0) + 2\sin(0)) = (0 + 2) - (-4 + 0) = 2 + 4 = 6$ Надеюсь, это поможет тебе в учёбе!

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