Задача 11890 Помогите пожалуйста..1,2 и з задачи...
Условие
Решение
1)=x^5/5|^2_(1)=(2^5-1^5)/5=31/5;
2)=sinx|^π_(0)=sinπ-sin0=0;
3)=x^4/4|^3_(1)=(3^4-1^4)/4=20;
4)=-ctgx|^(π/2)_(π/4)=-(0-1)=1;
5)=-1/(2*(2x+1))|^2_(1)=-(1/10)+(1/6)=1/15;
6)=6sin(x/2)^π_(0)=6sin(π/2)-0=6;
7)=-1/x|^(10)_(1)=(-1/10)+1=9/10;
8)=(-1/2)*cos2x|^(π/2)_(π/4)=-1/2cosπ+(1/2)cos(π/2)=1+0=1;
2.
1)слева: =tgx|^(π/4)_(0)=tg(π/4)-tg0=1
справа: =x|^(1)_(0)=1
слева 1 и справа 1.
Равенство верно
2)слева: =-cosx|^(π/3)_(0)=-cos(π/3)+cos0=-1/2+1=1/2
справа: =2sqrt(x)|^(1/4)_(1/16)=2*((1/2)-(1/4))=1/2
слева 1/2 и справа 1/2.
Равенство верно
3)слева: =sinx|^(π/2)_(0)=sin(π/2)-sin0=1
справа: =x^3/3|^(∛3)_(0)=3/3=1
слева 1 и справа 1.
Равенство верно
4)слева: =(2x+1)^2/4|^(1)_(0)=(9/4)-(1/4)=2
справа: =(x^4/4-x)|^(2)_(0)=((2^4/4)-2)=2
слева 2 и справа 2.
Равенство верно
3.
1)=-3сos(x/3)|^(2π)_(-π)=-3*cos(2π/3)+3cos(-π/3)=0;
2)=sqrt(2x+5)|^(2)_(-2)=sqrt(9)-sqrt(1)=3-1=2;^(3
3)=9tg(x/9)|^(3π)_(0)=9tg(π/3)-9tg0=3sqrt(3);
4)=2sqrt(x+3)|^(0)_(-2)=2sqrt(3)-2sqrt(1)=2sqrt(3)-1;
5)(sin(x/4)+cos(x/4))^2=1+2sin(x/2)
=(x-4cos(x/2))|^(2π/3)_(0)=(2π/3)-4cos(π/3)-0+4cos0=(2π/3)-4*(1/2)+4*1=(2π/3)+2.
6)=(1+2x^4)/8|^(2)_(0)=(33/8)-(1/8)=(32/8)=4.
7)=(x+(1/2)*sin2x)|^(π/12)_(0)=(π/12)+(1/2)*sin(π/6)=(π/12)+(1/4);
8)=((x^2/2)+2sqrt(x))|^(4)_(1)=(4^2/2)+2sqrt(4)-(1)^2/2-2sqrt(1)=
=8+4-(1/2)-2=9,5.