Question

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

Answer 1

We have, 
\[y = \left| x - 1 \right|\]
\[ \Rightarrow y = \begin{cases}x - 1&\text{ for }x \geq 1\\1 - x &\text{ for } x < 1\end{cases}\]
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)
\[y = 3 - \left| x \right|\]
\[ \Rightarrow y = \begin{cases}3 - x&\text{ for }x \geq o\\3 - \left( - x \right) = 3 + x&\text{ for }x < 0\end{cases}\]
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

\[y = x - 1\]
\[\text{ and }y = 3 - x\]
We get
\[ \Rightarrow x - 1 = 3 - x\]
\[ \Rightarrow 2x - 4 = 0\]
\[ \Rightarrow x = 2\]
\[ \Rightarrow y = 2 - 1 = 1\]
\[\text{ Thus P }\left( 2, 1 \right)\text{ is point of intersection of }y = x - 1\text{ and }y = 3 - x\]
Point of intersection for
\[y = 1 - x\]
\[y = 3 + x\]
\[ \Rightarrow 1 - x = 3 + x\]
\[ \Rightarrow 2x = - 2\]
\[ \Rightarrow x = - 1\]
\[ \Rightarrow y = 1 - \left( - 1 \right) = 2\]
\[\text{ Thus Q }\left( - 1, 2 \right)\text{ is point of intersection of }y = 1 - x\text{ and }y = 3 + x\]
\[\text{ Since the character of function changes at C }\left( 0, 3 \right)\text{ and A }(1, 0) ,\text{ draw AM perpendicular to }x - \text{ axis }\]
\[\text{ Required area = Shaded area }\left( QCPAQ \right)\]
\[ =\text{ Area }\left( QCB \right) +\text{ Area }\left( BCMAB \right) +\text{ area }\left( AMPA \right) . . . . . \left( 1 \right)\]
\[\text{ Area }\left( QCB \right) = \int_{- 1}^0 \left[ \left( 3 + x \right) - \left( 1 - x \right) \right]dx\]
\[ = \int_{- 1}^0 \left( 2 + 2x \right) dx\]
\[ = \left[ 2x + x^2 \right]_{- 1}^0 \]
\[ = 0 - \left( - 2 + 1 \right)\]
\[ = 1\text{ sq unit }. . . . . \left( 2 \right)\]
\[\text{ Area }\left( BCMA \right) = \int_0^1 \left[ \left( 3 - x \right) - \left( 1 - x \right) \right] dx\]
\[ = \int_0^1 2 dx \]
\[ = \left[ 2x \right]_0^1 = 2\text{ sq unit }. . . . . \left( 3 \right)\]
\[\text{ Area }\left( AMPA \right) = \int_1^2 \left[ \left( 3 - x \right) - \left( x - 1 \right) \right] dx\]
\[ = \int_1^2 \left( 4 - 2x \right) dx\]
\[ = \left[ 4x - x^2 \right]_1^2 \]
\[ = \left( 8 - 4 \right) - \left( 4 - 1 \right)\]
\[ = 1\text{ sq unit }. . . . . \left( 4 \right)\]
\[\text{ From }\left( 1 \right), \left( 2 \right), \left( 3 \right) \text{ and }\left( 4 \right)\]
\[\text{ Shaded area }= 1 + 2 + 1 = 4\text{ sq units }\]