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Q.
The area of the region bounded by the curves $y=|x-1|$ and $=3-|x|$ is
Application of Integrals
Solution:
Since, $y=|x-1|=\begin{cases} x-1, & x>1 \\ -x+1, & x \leq 1\end{cases}$
and
$y=3-|x|= \begin{cases}3+x, & x \leq 0 \\ 3-x, & x>0\end{cases}$
On solving $y=x-1$ and $y=3-x$, we get
$x-1=3-x$
$\Rightarrow x=2$
and $ y=3-2 \Rightarrow y=1$
Now, $ A B^2=(2-1)^2+(1-0)^2=1+1=2$
$\Rightarrow A B=\sqrt{2}$
and $ B C^2=(0-2)^2+(3-1)^2$
$=4+4=8$
$\Rightarrow B C=2 \sqrt{2}$
Area of rectangle $A B C D=A B \times B C$
$=\sqrt{2} \times 2 \sqrt{2}=4 sq$ units
Note The above approach is objective approach, for board (school) exams find area using integration.
