|2 (x^2+ 1 x^2 )+ | 1-x^2||=4 ( 3 2 -2^ x^2-3 - 1 2^ x^2+1 ) If x₁ and x₂ , where x₁ < x₂ , are two values of x satisfying the equation above, find the value of _ x₁-x₂ ^ 3 x₂-x₁ x 4 (1+ [ ( x 1+ x ) ] ) d x [Note: | | denotes the absolute value function, . denotes the fraction part function, [-] denotes the floor function]

Step-by-step solution

Since the equation is even on both sides,we only need to consider x d to find the positive solution x₂ and then x₁=-x₂ . For 0 x 1 , aligned 2 (x^2+ 1 x^2 )+1-x^2 & =4 ( 3 2 -2^ x^2-1 - 1 2^ x^2-1 ) x^2+ 2 x^2 +1 & =6-2^ x^2-1 - 1 2^ x^2-1 x^2+ 2 x^2 +2^ x^2-1 + 1 2^ x^2-1 & =5 aligned By inspection, the solution is x^2=1 or x₁=-1 and x^2=1 , then aligned I & = ₀^4 x 4 (1+ [ ( x 1+ x ) ] d x Note that [ ( x 1+ x ) ]=0 . & = ₀^4 x 4 d x= ₀^4 x 4 d x= . x^2 8 |₀ ^4=2 aligned