Question

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

Answer 1

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