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Question : 75 of 160
Marks:
+1,
-0
Solution:
We have,
‌[sin‌+sin‌+...+sin‌] Here,
r th term
=sin(‌) ∴ The given limit,
‌‌sin(‌) Put
‌→x,‌→dx,Σ→∫ xmin=‌=0 xmax=‌=‌=sin‌x‌d‌x =(−cos‌x)0π∕2=−(cos‌x)0π2 ‌‌=−[cos‌−cos‌0]=−[0−1]=1
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