A) \[\frac{1}{2}\]
B) 1
C) \[\frac{3}{2}\]
D) 2
Correct Answer: B
Solution :
[b] \[\left| x \right|\] for \[x\ge 0\] = x and \[\left| x-1 \right|\]for \[x\le 1=-(x-1),\] So, \[\int_{0}^{1}{(\left| x \right|+\left| x-11 \right|)=}\] required area \[a=\int_{0}^{1}{xdx-\int_{0}^{1}{(x-1)dx}}\] \[=\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}-\left[ \frac{{{x}^{2}}}{2}-x \right]_{0}^{1}=\frac{1}{2}-\left( \frac{1}{2}-1 \right)=1\] sq. units
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