Question

Find the area of the region bounded by y = | x − 1 | and y = 1.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=\left|x-1\right| \\ \Rightarrow y=\left\{\begin{array}{l}x-1x\ge 1\\ 1-xx<1\end{array}\right. \end{gathered}$$
y = x − 1 is a straight line originating from A(1, 0) and making an angle 45^o with the x-axis
y = 1 − x is a straight line originating from A(1, 0) and making an angle 135^o with the x-axis

y = x is a straight line parallel to x-axis and passing through B(0, 1)

The point of intersection of two lines with y = 1 is obtained by solving the simultaneous equations

$$\begin{gathered} y=1 \\ \mathrm{and}y=x-1 \\ \Rightarrow 1=x-1 \\ \Rightarrow x-2=0 \\ \Rightarrow x=2 \\ \Rightarrow C\left(2,1\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}y=x-1\mathrm{and}y=1 \\ y=1\mathrm{and}y=1-x \\ \Rightarrow 1=1-x \\ \Rightarrow x=0 \\ \Rightarrow B\left(0,1\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}y=1-x\mathrm{and}y=1 \\ \mathrm{Since}y=\left|x-1\right|\mathrm{changes}\mathrm{character}\mathrm{at}A(1,0),\mathrm{Consider}\mathrm{point}\mathrm{P}(1,1)\mathrm{on}\mathrm{BC}\mathrm{such}\mathrm{that}\mathrm{PA}\mathrm{is}\mathrm{perpendicular}\mathrm{to}x\mathit{-}\mathrm{axis}. \\ \mathrm{Required}\mathrm{shaded}\mathrm{area}\left(\mathrm{ABCA}\right)=\mathrm{area}\left(ABPA\right)+\mathrm{area}\left(PCAP\right) \\ ={\int }_0^1\left[1-\left(1-x\right)\right]dx+{\int }_1^2\left[1-\left(x-1\right)\right]dx \\ ={\int }_0^{1}xdx+{\int }_1^2\left(2-x\right)dx \\ ={\left[\frac{x^2}{2}\right]}_0^1+{\left[2x-\frac{x^2}{2}\right]}_1^2 \\ =\frac{1}{2}+\left[4-2-2+\frac{1}{2}\right] \\ =\frac{1}{2}+\frac{1}{2}=1\mathrm{sq}.\mathrm{unit} \end{gathered}$$