A) 2
B) 1
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
\[y=x-1,\]if \[x>1\]and \[y=-(x-1),\]if \[x<1\] Area \[=\int_{0}^{1}{(1-x)dx+\int_{1}^{2}{(x-1)dx}}=\left[ x-\frac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\] \[=\left[ 1-\frac{1}{2} \right]+\left[ -\left( \frac{1}{2}-1 \right) \right]\]\[=\frac{1}{2}+\frac{1}{2}=1\].You need to login to perform this action.
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