question_answer

The area of the region bounded by the curves \[y=\left| x-1 \right|\] and \[y=3-\left| x \right|\] is

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