ज्ञात करे [ Evaluate] `int_(1)^(4)f(x)dx, "` जहाँ (where) ` f(x) = |x -1| + |+| x-2| + | x- 3|`

Explanation

यहाँ `|x-1|,|x-2|` तथा`|x-3|` आता है
अब , ` x-1 = 0 rArr x = 1 , x - 2 = 0 rArr x = 2 " तथा " x - 3 = 0 rArr x = 3 `
` (##KCS_HIN_MAT_XII_P2_C21_SLV_005_S01.png" width="80%">
अब, ` = int _(1)^(4) f(x) dx = int_(1)^(2) f(x) dx + int _(2)^(3)f(x) dx + int _(3)^(4) f(x) dx `
= ` int _(1)^(2) [{ (x-1)- (x-2) - (x-3) } dx + int_(2)^(3) {(x-1)+(x-2) - (x-3) }] dx `
` + int _(3)^(4) {(x-1)+(x-2)+(x-3)dx}`
`= int _(1)^(2) (-x+4) dx + int _(2)^(3) x dx + int _(3)^(4) (3x-6)dx `
` = [ (-x^(2))/2 + 4x]_(1)^(2) +[ (x^(2))/2] _(2)^(3) + [ (3x^2)/2-6x]_(3)^(4) = ( 5/2 +5/2 + 9/2 )=19/2`