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