Задача 35234 Решите пожалуйста ...

nastyaaaa

2019-04-02 14:38:37

Решите пожалуйста

sova

2019-04-02 16:00:24

∫ ^(3)_(0,5)sqrt(2x+3)dx=(1/2) ∫ ^(3)_(0,5)(2x+3)^(1/2)d(2x+3)=

=(2x+3)^((1/2)+1)/((1/2)+1)=(2/3)(2x+3)^(3/2)|^(3)_(0,5)=

=(2/3)*(2*3+3)^(3/2) - (2/3)*(2*0,5+3)^(3/2)=

=(2/3)3^3-(2/3)*2^3=(2/3)*(27-8)= [b]38/3[/b]


∫ ^(1)_(-2)(3x^2-4x-1)dx= (x^3-2x^2-x)|^(1)_(-2)=

=(1-2-1)-(-8-8-(-2))= [b]...[/b]



∫ ^(π)_(0)cos^2xsinxdx=∫ ^(π)_(0)cos^2x(-d(cosx))=

= -(cos^3x/3)|^(π)_(0)=-(1/3)*(cos^(3)(π)-cos0)=(-1/3)*(-1-1)= [b]2/3[/b]


∫ ^(1)_(-1)(4xS+1)^3dx=(1/(4S)) ∫ ^(1)_(-1)(4xS+1)^3d(4xS+1)=

=(1/(4S)) *(4xS+1)^4/4|^(1)_(-1)= [b](1/(16S)) *((4S+1)^4-(1-4S)^4)[/b]