`int_(0)^(1)|3x-1|dx=`

int_(0)^(1)3^(x)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_(0)^(1)[3x]dx

Evaluate: int_0^1 |3x-1| dx

int_(0)^(1) |5x -3 | dx =?

int_(0)^(1) |5x -3 | dx =?

int_(0)^(1)x^(3)dx

int_(0)^(3)(x+1)dx

int_(0)^(1)log(3+x)dx

l_(1)=int_(0)^(1)3^(x^(2))dx,l_(2)=int_(0)^(1)3^(x^(3))dx,l_(3)=int_(1)^(2)3^(x^(2))dx,l_(4)=int_(1)^(2)3^(x^(3))dx then (i)I_(1)>I_(2)(ii)I_(2)>I_(1)(iii)I_(3)>I_(4)(iv)I_(4)>I_(3)