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Question : 79 of 160
Marks:
+1,
-0
Solution:
‌sin[x−[x]]‌d‌x =−1sin(x+2)‌d‌x+sin(x+1)‌d‌x+sin(x)‌d‌x +sin(x−1)‌d‌x =cos(2−‌)−cos(1)+cos‌0−cos‌1+cos‌0−cos‌1 =3(1−cos‌1)+sin‌2−sin‌1+cos‌0−cos(‌−1)
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