A) 6 sq. units
B) 2 sq. units
C) 9 sq. units
D) 4 sq. units.
Correct Answer: D
Solution :
 \[A=\int\limits_{-1}^{0}{\left\{ \left( 3+x \right)-\left( -x+1 \right) \right\}dx+}\]\[\int\limits_{0}^{1}{\left\{ \left( 3-x \right)-\left( -x+1 \right) \right\}dx+}\]\[\int\limits_{1}^{2}{\left\{ \left( 3-x \right)-\left( x-1 \right) \right\}dx}\] |
| \[=\int\limits_{-1}^{0}{\left( 2+2x \right)dx+\int\limits_{0}^{1}{2dx+\int\limits_{1}^{2}{\left( 4-2x \right)dx}}}\] |
| \[=[2x-{{x}^{2}}]_{-1}^{0}+[2x]_{\,\,\,0}^{1}+{{[4x-{{x}^{2}}]}_{1}}^{2}\] |
| \[=0-\left( -2+1 \right)+\left( 2-0 \right)+\left( 8-4 \right)-\left( 4-1 \right)\] |
| \[=1+2+4-3=4sq.units\] |
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