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Question : 75 of 160
Marks:
+1,
-0
Solution:
Let
I=‌‌dx =‌‌dx Let
x−2=t⇒‌‌dx=dt When
x→0,t→−2 and
x→4,t→2 I‌=‌‌dt‌=‌‌‌‌dt+4‌‌‌dt‌=−[√4−t2]−22+4[sin‌−1‌]−22‌=0+4[sin‌−11−sin‌−1(−1)]‌=0+4[‌+‌]=4π
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