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B
30
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C
29
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D
28
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Solution
The correct option is C 29 ∫5−5|x+2|dx ∫−2−5−(x+2)dx+∫5−2(x+2)dx =−∣∣∣x22∣∣∣−2−5−2|x|−2−5+∣∣∣x22∣∣∣5−2+2|x|5−2 −[42−252]−2[−2+5]+[252−42]+2[5+2] =212−6+212+14 =21+8 =29