₀^1 x |x- 1 2 | d x=

Step-by-step solution

The integrand x |x- 1 2 | has a discontinuity in its derivative at x = 1 2 , requiring integration over two intervals. For x ₀^1 x |x- 1 2 | dx = ₀^ 1/2 x ( 1 2 - x ) dx + _ 1/2 ^1 x (x - 1 2 ) dx The first integral evaluates as: ₀^ 1/2 ( x 2 - x^2 ) dx = [ x^2 4 - x^3 3 ]₀^ 1/2 = 1 16 - 1 24 = 1 48 The second integral yields: _ 1/2 ^1 (x^2 - x 2 ) dx = [ x^3 3 - x^2 4 ]_ 1/2 ^1 = ( 1 3 - 1 4 ) - ( 1 24 - 1 16 ) = 1 12 + 1 48 = 5 48 Summing both parts gives 1 48 + 5 48 = 6 48 = 1 8 . Final result: 1 8