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Question : 79 of 160
Marks:
+1,
-0
Solution:
Let‌I=∫−{[x]+log()‌d‌x} =∫−‌‌[x]dx+∫−log()‌d‌x Let
f(x)=log() f(−x)=log()=−f(x) ‌‌‌‌ [odd function]
=∫−0‌(−1)dx+∫0‌0‌dx =−(x)−0+0=−(0+)=−
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