`int_(0)^(4)|x-1|dx`

Explanation

माना `I=int_(0)^(4)|x-1|dx`
`therefore" "I=int_(0)^(1)|x-1|dx+int_(1)^(4)|x-1|dx`
`=int_(0)^(1)(1-x)dx+int_(1)^(4)(x-1)dx`
`=[x-(x^(2))/(2)]_(0)^(1)+[(x^(2))/(2)-x]_(1)^(4)`
`=(1-(1)/(2))-0+((4^(2))/(2)-4)-((1)/(2)-1)`
`=(1)/(2)+4+(1)/(2)=5`