Question

Find the area of the figure bounded by the curves y = | x − 1 | and y = 3 −| x |.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=\left|x-1\right| \\ \Rightarrow y=\left\{\begin{array}{l}x-1\mathrm{for}x\ge 1\\ 1-x\mathrm{for}x<1\end{array}\right. \end{gathered}$$
y = x − 1 is a straight line passing through A(1, 0)
y = 1 − x is straight line passing through A(1, 0) and cutting y-axis at B(0, 1)

$$\begin{gathered} y=3-\left|x\right| \\ \Rightarrow y=\left\{\begin{array}{l}3-x\mathrm{for}x\ge o\\ 3-\left(-x\right)=3+x\mathrm{for}x<0\end{array}\right. \end{gathered}$$

y = 3 − x is straight line passing through C(0, 3) and D(3, 0)
y = 3 + x is a straight line passing through C(0, 3) and D'(−3, 0)
The point of intersection is obtained by solving the simultaneous equations
$$\begin{gathered} y=x-1 \\ \mathrm{and}y=3-x \\ \mathrm{We}\mathrm{get} \\ \Rightarrow x-1=3-x \\ \Rightarrow 2x-4=0 \\ \Rightarrow x=2 \\ \Rightarrow y=2-1=1 \\ \mathrm{Thus}P\left(2,1\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}y=x-1\mathrm{and}y=3-x \\ \mathrm{Point}\mathrm{of}\mathrm{intersection}\mathrm{for} \\ y=1-x \\ y=3+x \\ \Rightarrow 1-x=3+x \\ \Rightarrow 2x=-2 \\ \Rightarrow x=-1 \\ \Rightarrow y=1-\left(-1\right)=2 \\ \mathrm{Thus}\mathrm{Q}\left(-1,2\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}y=1-x\mathrm{and}y=3+x \\ \mathrm{Since}\mathrm{the}\mathrm{character}\mathrm{of}\mathrm{function}\mathrm{changes}\mathrm{at}C\left(0,3\right)\mathrm{and}A(1,0),\mathrm{draw}AMp\mathrm{erpendicular}\mathrm{to}x\mathit{-}\mathrm{axis} \\ \mathrm{Required}\mathrm{area}=\mathrm{Shaded}\mathrm{area}\left(QCPAQ\right) \\ =\mathrm{Area}\left(QCB\right)+\mathrm{Area}\left(BCMAB\right)+\mathrm{area}\left(AMPA\right).....\left(1\right) \\ \mathrm{Area}\left(QCB\right)={\int }_{-1}^0\left[\left(3+x\right)-\left(1-x\right)\right]dx \\ ={\int }_{-1}^0\left(2+2x\right)dx \\ ={\left[2x+x^2\right]}_{-1}^0 \\ =0-\left(-2+1\right) \\ =1\mathrm{sq}\mathrm{unit}.....\left(2\right) \\ \mathrm{Area}\left(BCMA\right)={\int }_0^1\left[\left(3-x\right)-\left(1-x\right)\right]dx \\ ={\int }_0^{1}2dx \\ ={\left[2x\right]}_0^1=2\mathrm{sq}\mathrm{unit}.....\left(3\right) \\ \mathrm{Area}\left(AMPA\right)={\int }_1^2\left[\left(3-x\right)-\left(x-1\right)\right]dx \\ ={\int }_1^2\left(4-2x\right)dx \\ ={\left[4x-x^2\right]}_1^2 \\ =\left(8-4\right)-\left(4-1\right) \\ =1\mathrm{sq}\mathrm{unit}.....\left(4\right) \\ \mathrm{From}\left(1\right),\left(2\right),\left(3\right)\mathrm{and}\left(4\right) \\ \mathrm{Shaded}\mathrm{area}=1+2+1=4\mathrm{sq}\mathrm{units} \end{gathered}$$