Question

Find the area enclosed by the curve $y = |x - 1|\,\,and\,\,y = - |x - 1| + 1$ .

Answer 1

Hint**:** To find the area enclosed between two curves. First we have to draw a rough sketch of given curves and then from the sketch we find the common region and using the given curve we find their point of intersection of common regions and hence by using the concept of definite integral we find enclosed areas between given curves.

Complete step-by-step answer :

The curves are

$y = |x - 1|$ and $y = - |x - 1| + 1$

First we will make a table for putting the different value of $x$ in both equation

The two given curves are

I. $y = |x - 1|$

II. $y = - |x - 1| + 1$

let's consider the first curve,

I. $y = |x - 1|$

If $x = 0$ therefore

$y = |0 - 1|$

$y = 1$

If $x = 0.5$

$y = |0.5 - 1|$

$y = | - 0.5|$

$y = 0.5$

$if\,\,x = 1$

$y = |1 - 1|$

$y = 0$

If $x = 1.5$

$y = |1.5 - 1|$

$y = |0.5|$

x	$0$	$0.5$	$1$	$1.5$
y	$1$	$0.5$	$0$	$0.5$

II. $y = - |x - 1| + 1$

$if\,\,x = 0$

$y = - |0 - 1| + 1$

$y = - 1 + 1$

$y = 0$

$if\,x = 0.5$

$y = - |0.5-1| + 1$

$y = - | - 0.5| + 1$

$y = - 0.5 + 1$

$y = 0.5$

$if\,\,x = 1$

$then\,\,y = |1 - 1| + 1$

$y = 0 + 1$

$y = 1$

If $x = 1.5$

$y = -|x - 1| + 1$

$y = -|1.5 - 1| + 1$

$y = -|0.5| + 1$

$y =- 0.5 + 1$

$y = 0.5$

x	$0$	$0.5$	$1$	$1.5$
y	$0$	$0.5$	$1$	$0.5$

Now, we represent these points on $x,y$ plane, as:

Here, $PQRS$ is a square. And the coordinates are $P(1,0),Q(1.5,0.5),\,R(1,1)\,\,and\,\,S(0.5,0.5)$

Now, we will calculate the length of sides by using distance formula, we have

In the coordinates $P(1,0)$ and $Q(1.5,0.5)$ we use distance formula to calculate length of sides so,

$\Rightarrow PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$

$= \sqrt {{{(1.5 - 1)}^2} + {{(0.5)}^2}}$

$= \sqrt {2 \times {{(0.5)}^2}}$

$PQ = \sqrt {0.5}$

Now, we will calculate the area of the square. As we know that all sides are equal in length.

Area of the square $=$ side$\times$side

Then, we will put the value of side in the formula, so

$\Rightarrow$ Area of square $= \sqrt {0.50} \times \sqrt {0.50}$

Area of square $= {\left( {\sqrt {0.50} } \right)^2}$

Area of square $= 0.50$ square units

Hence, the area enclosed between the curves is $0.50$ square units.

Note**:** In case of mode problems we must find critical points. Because, we know that the left area of critical points is negative and the right area is positive and therefore the sketch of curves having mode always has two parts with inclination of ${45^0}$ . From curves we first find points of intersection and use these to find the area enclosed between them by using definite integral.