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Question : 78 of 160
Marks:
+1,
-0
Solution:
‌I=|(2−x2)|‌dx‌=(x2−2)‌dx++√2(2−x2)‌dx++4(x2−2)‌dx‌=[‌−2x]−2−√2+[2x−‌]−√2√2+[‌−2x]√24‌⇒‌+2√2)−(‌+4)‌+[(2√2−‌)−(−2√2+‌)]‌+[(‌−8)−(‌−2√2)]‌⇒‌−‌+4√2−‌+‌+‌‌=12+‌
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