" 3.3."int_(1)^(4)[1x-11+|x-2|+|x-3|]dx

int_(0)^(4)(|x-1|+|x-2|+|x-3|)dx=

int_(1)^(3)(|x-1|)/(|x-2|+|x-3|)dx=

Evaluate: int_(0)^(1)|5x-3|dx( ii) int_(0)^( pi)|cos x|dx( iii) int_(-5)^(5)|x-2|dx( iv )int_(-1)^(1)e^(|x|)dx(v)int_(0)^(2)|x^(2)+2x-3|dx(v)int_(1)^(4)(|x-1|+|x-2|+|x-3|)dx( vi) int_(1)^(2)|x^(3)-x|dx

int_(1)^(3)(x-1)(x-2)(x-3)dx=

int_(1)^(3)(x-1)(x-2)(x-3)dx=

int_(1)^(3)(x-1)(x-2)(x-3)dx=

(11) int_(2)^(4)|x-1|dx

int_(1)^(3)(3x-4)*dx

int_(1)^(4)3x^(2)dx

int_(1)^(3)(3x-4)dx