3. Вычислите следующие интегралы:

Решение:

a) \( \int_{1}^{2} 2x^3 dx \)

\( = 2 \cdot \frac{x^4}{4} \Big|_{1}^{2} = \frac{1}{2} x^4 \Big|_{1}^{2} = \frac{1}{2} (2^4 - 1^4) = \frac{1}{2} (16 - 1) = \frac{15}{2} \)

б) \( \int_{1}^{4} \sqrt{2x} dx \)

\( = \int_{1}^{4} \sqrt{2} \sqrt{x} dx = \sqrt{2} \int_{1}^{4} x^{1/2} dx = \sqrt{2} \cdot \frac{x^{3/2}}{3/2} \Big|_{1}^{4} = \sqrt{2} \cdot \frac{2}{3} x^{3/2} \Big|_{1}^{4} = \frac{2\sqrt{2}}{3} (4^{3/2} - 1^{3/2}) = \frac{2\sqrt{2}}{3} (8 - 1) = \frac{14\sqrt{2}}{3} \)

в) \( \int_{1}^{e} \frac{1}{x} dx \)

\( = \ln|x| \Big|_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1 \)

г) \( \int_{0}^{\pi/2} \sin 2x dx \)

\( = -\frac{\cos 2x}{2} \Big|_{0}^{\pi/2} = -\frac{1}{2} (\cos(\pi) - \cos(0)) = -\frac{1}{2} (-1 - 1) = -\frac{1}{2} (-2) = 1 \)

д) \( \int_{1}^{4} \frac{1}{\sqrt{x}} dx \)

\( = \int_{1}^{4} x^{-1/2} dx = \frac{x^{1/2}}{1/2} \Big|_{1}^{4} = 2 \sqrt{x} \Big|_{1}^{4} = 2 (\sqrt{4} - \sqrt{1}) = 2 (2 - 1) = 2 \)

Ответ: a) \( \frac{15}{2} \); б) \( \frac{14\sqrt{2}}{3} \); в) 1; г) 1; д) 2.