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Question : 11 of 150
Marks:
+1,
-0
Solution:
Let
I={|sin‌x|−|‌sin‌x|}dx ={|sin‌x|−|sin‌x|}dx =|sinx|dx =[|sin‌x|dx+|sin‌x|dx+...... +|sin‌x|dx] Now,
I1=|sinx|dx I1=sin‌x‌d‌x−sin‌x‌d‌x =[−cos‌x]0π+[cos‌x]π2π =−[−1−1]+[+1+1] =2+2 ∴I=[4+4+4+......n times ] =(4n)=2n=4
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