Let _ -2 ² (| x| + [x x]) ,dx = 2(3 - 2) + , where [ ] is the greatest integer function. Then ( 2 ) equals:

Step-by-step solution

Let I = _ -2 ² (| x| + [x x] ) dx = I₁ + I₂ , where I₁ = _ -2 ² | x| , dx and I₂ = _ -2 ² [x x] , dx . Evaluating I₁ : Since | x| is even, I₁ = 2 ₀² | x| , dx . For x [0, 2] , 2 I₁ = 2 ₀² x , dx = 2 [- x ]₀^2 = 2(1 - 2) = 2 - 2 2 Evaluating I₂ : Let g(x) = x x . Then g(-x) = (-x) (-x) = x x = g(x) , so g(x) is even, and hence [x x] is also even. I₂ = 2 ₀² [x x] , dx Behaviour of y = x x on [0, 2] : At x = 0 , y = 0 . At x = 2 , y = 2 2 1.818 . y' = x + x x > 0 on [0, 2] , so y is strictly increasing on this interva