A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
Required area \[=\int_{1}^{3}{y}\,dx\] \[=\int_{1}^{3}{|x-2|}dx\] \[=\int_{1}^{2}{-(x-2)dx}+\int_{2}^{3}{(x-2)}dx\] \[=\int_{1}^{2}{(2-x)dx}+\int_{2}^{3}{(x-2)}dx\] \[=\left[ 2x-\frac{{{x}^{2}}}{2} \right]_{1}^{2}+\left[ \frac{{{x}^{2}}}{2}-2x \right]_{2}^{3}\] \[=(4-2)-\left( 2-\frac{1}{2} \right)+\left( \frac{9}{2}-6 \right)-(2-4)\] \[=2-\frac{3}{2}-\frac{3}{2}+2\] \[=4-3=1\]
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