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I am trying to solve this problem and I can’t. I found the areas of each part, starting from the left to right I got -1, -2pi, and 4. The area of the same definite integral but without the absolute value sign is -2pi+3, so I don’t understand why the answer couldn’t be the absolute value of -2pi+3. The answer in the picture (1/2) I got from using the fundamental theorem of Calc, with f’(-4) = 1/2 and f’(6)= -1 then f’(6)-f’(-4).
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Just reflect the negative part across the x axis and sum the areas.
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