5. Вычислите интегралы: a) ∫₀² (3x² - 2x + 4)dx; б) ∫_{1/2}¹ (2/x³ + 8)dx; в) ∫₃⁶ (4x - 1/(2√x-2))dx; г) ∫₀^{π/6} 3sin 3xdx.

Решение:

a) \( \int_{0}^{2} (3x^2 - 2x + 4) dx \)

Первообразная: \( F(x) = x^3 - x^2 + 4x \)

\( [x^3 - x^2 + 4x]_{0}^{2} = (2^3 - 2^2 + 4 \cdot 2) - (0) = (8 - 4 + 8) = 12 \)

б) \( \int_{1/2}^{1} (\frac{2}{x^3} + 8) dx = \int_{1/2}^{1} (2x^{-3} + 8) dx \)

Первообразная: \( F(x) = 2\frac{x^{-2}}{-2} + 8x = -x^{-2} + 8x = -\frac{1}{x^2} + 8x \)

\( [-\frac{1}{x^2} + 8x]_{1/2}^{1} = (-\frac{1}{1^2} + 8 \cdot 1) - (-\frac{1}{(1/2)^2} + 8 \cdot \frac{1}{2}) \)

\( = (-1 + 8) - (-\frac{1}{1/4} + 4) = 7 - (-4 + 4) = 7 - 0 = 7 \)

в) \( \int_{3}^{6} (4x - \frac{1}{2\sqrt{x-2}}) dx \)

Первообразная: \( F(x) = \int 4x dx - \int \frac{1}{2} (x-2)^{-\frac{1}{2}} dx = 2x^2 - \frac{1}{2} \frac{(x-2)^{\frac{1}{2}}}{\frac{1}{2}} = 2x^2 - \sqrt{x-2} \)

\( [2x^2 - \sqrt{x-2}]_{3}^{6} = (2 \cdot 6^2 - \sqrt{6-2}) - (2 \cdot 3^2 - \sqrt{3-2}) \)

\( = (2 \cdot 36 - \sqrt{4}) - (2 \cdot 9 - \sqrt{1}) = (72 - 2) - (18 - 1) = 70 - 17 = 53 \)

г) \( \int_{0}^{\frac{\pi}{6}} 3\sin(3x) dx \)

Первообразная: \( F(x) = 3 \frac{-\cos(3x)}{3} = -\cos(3x) \)

\( [-\cos(3x)]_{0}^{\frac{\pi}{6}} = -\cos(3 \cdot \frac{\pi}{6}) - (-\cos(3 \cdot 0)) \)

\( = -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 - (-1) = 1 \)

Ответ: a) 12; б) 7; в) 53; г) 1.