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Question : 77 of 160
Marks:
+1,
-0
Solution:
From given expression,
I=|x2−3x+2|dxAs,
x2−3x+1=(x−2)(x−1)x2−3x+1<0,x∈(1,2)And
x2−3x+2≥0,x∈R−(1,2)So,
I=[(x2−3x+2)dx−(x2−3x+2)dx +(x2−3x+2)dx]=
[[−+2x]01−[−+2x]12 +[−+2x]23]=
[(−+2)−[(−+4) −(−+2)]+[(−+6) −(−+4)]]=
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