5. Вычислите интегралы: a) ∫₀³ (x² + 4x - 1)dx; б) ∫₁³ (3 - 1/x²)dx; в) ∫₀³ (2/√x+1 + 3x²)dx; г) ∫₀^{π/2} 1/2 cos x/2 dx.

Решение:

a) \( \int_{0}^{3} (x^2 + 4x - 1) dx \)

Первообразная: \( F(x) = \frac{x^3}{3} + 2x^2 - x \)

\( [\frac{x^3}{3} + 2x^2 - x]_{0}^{3} = (\frac{3^3}{3} + 2 \cdot 3^2 - 3) - (0) = (\frac{27}{3} + 2 \cdot 9 - 3) = (9 + 18 - 3) = 24 \)

б) \( \int_{1}^{3} (3 - \frac{1}{x^2}) dx \)

Первообразная: \( F(x) = 3x - \frac{x^{-1}}{-1} = 3x + \frac{1}{x} \)

\( [3x + \frac{1}{x}]_{1}^{3} = (3 \cdot 3 + \frac{1}{3}) - (3 \cdot 1 + \frac{1}{1}) = (9 + \frac{1}{3}) - (3 + 1) = 9 + \frac{1}{3} - 4 = 5 + \frac{1}{3} = \frac{15+1}{3} = \frac{16}{3} \)

в) \( \int_{0}^{3} (\frac{2}{\sqrt{x+1}} + 3x^2) dx \)

Первообразная: \( F(x) = \int 2(x+1)^{-\frac{1}{2}} dx + \int 3x^2 dx = 2 \frac{(x+1)^{\frac{1}{2}}}{\frac{1}{2}} + x^3 = 4\sqrt{x+1} + x^3 \)

\( [4\sqrt{x+1} + x^3]_{0}^{3} = (4\sqrt{3+1} + 3^3) - (4\sqrt{0+1} + 0^3) = (4\sqrt{4} + 27) - (4\sqrt{1}) = (4 \cdot 2 + 27) - 4 = (8 + 27) - 4 = 35 - 4 = 31 \)

г) \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cos(\frac{x}{2}) dx \)

Первообразная: \( F(x) = \frac{1}{2} \frac{\sin(\frac{x}{2})}{\frac{1}{2}} = \sin(\frac{x}{2}) \)

\( [\sin(\frac{x}{2})]_{0}^{\frac{\pi}{2}} = \sin(\frac{\pi/2}{2}) - \sin(\frac{0}{2}) = \sin(\frac{\pi}{4}) - \sin(0) = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2} \)

Ответ: a) 24; б) \( \frac{16}{3} \); в) 31; г) \( \frac{\sqrt{2}}{2} \).