u=3x
du=3dx
dx=(1/3)du
1.=(1/3) ∫ ^(1/3)_(0)d(3x)/sqrt(1-(3x)^2)=
=(1/3)arcisn(3x)|^(1/3)_(0)=(1/3)arcsin1-(1/3)arcsin0=(1/3)*(π/2)-0=
= [b]π/6[/b]
2.
∫ sqrt(u)du= ∫ u^(1/2)du=u^(3/2)/(3/2)+C=(2/3)(u^(3/2)+C
Поэтому
∫ ^(6)_(2)sqrt(x-2)d(x-2)=(2/3)*(x-2)^(3/2)|^(6)_(2)=
=(2/3)(4^(3/2))-(2/3)(0^(3/2))=(2/3)*8= [b]16/3[/b]