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Question

Evaluate
ba|x|xdx

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Solution

I=ba|x|xdx

case 1 : b>a>0

I=ba|x|xdx=baxxdx=badx=ba=|b||a|

case 2 :b>0>a

I=ba|x|xdx=0axxdx+b0xxdx=0adx+b0dx=|b||a|

Case 3: 0>b>a

I=ba|x|xdx=baxxdx=badx=(ba)=|b||a|

I=ba|x|xdx=|b||a|.

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