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Question

The value of 21 f(x) dx, where f(x)=(x+1|+(x|+(x1| is

A
92
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B
192
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C
152
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D
212
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Solution

The correct option is B 192
We can redefine f as
f(x)= 2x, if 1<x0x+2, if 0<x1 3x, if 1<x221f(x) dx=01 f(x) dx+10f(x) dx+21f(x) dx =01(2x) dx+10(x+2) dx+213x dx =(2xx22)01+(x22+2x)10 +3(x22)21=(2+12)+(12+2)+3(212)=192

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