[m]= (\frac{5 \cdot 2^3}{3} + 3 \cdot 2^2 - 2) - (\frac{5 \cdot (-1)^3}{3} + 3 \cdot (-1)^2 - (-1))= [/m]
[m]=(\frac{40}{3} + 12 - 2) - (-\frac{5}{3} + 3 + 1) = \frac{45}{3} + 10 - 4 = 15 + 6 = 21[/m]
б) [m]\int \limits_{0}^{\pi/12} \sin 3x\ dx = \frac{1}{3} (-\cos 3x)\ |_{0}^{\pi/12} = -\frac{1}{3} (\cos \frac{\pi}{4} - \cos 0) = [/m]
[m]= -\frac{1}{3}(\frac{\sqrt{2}}{2} - 1) = \frac{1}{3} \cdot (1 - \frac{\sqrt{2}}{2})= \frac{1}{3} \cdot \frac{2-\sqrt{2}}{2} = \frac{2-\sqrt{2}}{6}[/m]