Question

Find the area enclosed by the curves y = | x − 1 | and y = −| x − 1 | + 1.

Answer 1

The given curves are
$y=\left|x-1\right|.....\left(1\right)$
$y=-\left|x-1\right|+1.....\left(2\right)$
Clearly $y=\left|x-1\right|$ is cutting the x-axis at (1, 0) and the y-axis at (0, 1) respectively.
Also $y=-\left|x-1\right|+1$ is cutting both the axes at (0, 0) and x-axis at (2, 0).
We have,
$$\begin{gathered} y=\left|x-1\right| \\ y=\left\{\begin{array}{l}x-1x\ge 1\\ 1-xx<1\end{array}\right. \\ \mathrm{And} \\ y=-\left|x-1\right|+1 \\ y=\left\{\begin{array}{l}2-xx\ge 1\\ xx<1\end{array}\right. \\ \mathrm{Solving}\mathrm{both}\mathrm{the}\mathrm{equations}\mathrm{for}x<1 \\ y=1-x\mathrm{and}y=x, \\ \mathrm{We}\mathrm{get}x=\frac{1}{2}\mathrm{and}y=\frac{1}{2} \\ \mathrm{And}\mathrm{solving}\mathrm{both}\mathrm{the}\mathrm{equations}\mathrm{for}x\ge 1 \\ y=x-1\mathrm{and}y=2-x, \\ \mathrm{We}\mathrm{get}x=\frac{3}{2}\mathrm{and}y=\frac{1}{2} \end{gathered}$$
Thus the intersecting points are $\left(\frac{1}{2},\frac{1}{2}\right)$ and $\left(\frac{3}{2},\frac{1}{2}\right)$.
The required area A = ( Area of ABFA + Area of BCFB)
Now approximating the area of ABFA the length = $\left|y_1\right|$ and width = dx
Area of ABFA
$$\begin{gathered} ={\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\left[x-\left(1-x\right)\right]\mathit{d}x \\ ={\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\left(2x-1\right)\mathit{d}x \\ ={\left[x^2-x\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1 \\ =\frac{1}{4} \end{gathered}$$
Similarly approximating the area of BCFB the length $=\left|y_2\right|$ and width= dx
Area of BCFB
$$\begin{gathered} ={\int }_1^{\frac{3}{2}}\left[\left(2-x\right)-\left(x-1\right)\right]\mathit{d}x \\ ={\int }_1^{\frac{3}{2}}\left(3-2x\right)\mathit{d}x \\ ={\left[3x-x^2\right]}_1^{\frac{3}{2}} \\ =\frac{1}{4} \end{gathered}$$
Thus the required area A =( Area of ABFA + Area of BCFB)
$=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
Hence the required area is $\frac{1}{2}$ square units.