_ -2 ^2 |x-3| x-3 d x=

Step-by-step solution

The integrand is f(x) = |x-3| x-3 . Recall the definition of the absolute value function: |x-3| = x-3 if x-3 > 0 (i.e., x > 3 ) and |x-3| = -(x-3) if x-3 For the interval [-2, 2] , we have x Therefore, for all x [-2, 2] , |x-3| = -(x-3) . Substituting this into the integrand, we get f(x) = -(x-3) x-3 = -1 . The integral becomes _ -2 ² (-1) dx . Evaluating the definite integral: [-x]_ -2 ² = -(2) - (-(-2)) = -2 - 2 = -4 . Answer: -4