Question
The area of the region bounded by the curves $$y = \left| {x - 1} \right|$$ and $$y = 3 - \left| x \right|$$ is-
A.
$$6$$ square units
B.
$$2$$ square units
C.
$$3$$ square units
D.
$$4$$ square units
Answer :
$$4$$ square units
Solution :
$$\eqalign{ & A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} \cr & = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + } \int\limits_0^1 {2\,dx + } \int\limits_1^2 {\left( {4 - 2x} \right)dx} \cr & = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2 \cr & = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right) \cr & = 1 + 2 + 4 - 3 \cr & = 4\,{\text{sq}}{\text{.}}\,{\text{units}} \cr} $$
$$\eqalign{ & A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} \cr & = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + } \int\limits_0^1 {2\,dx + } \int\limits_1^2 {\left( {4 - 2x} \right)dx} \cr & = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2 \cr & = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right) \cr & = 1 + 2 + 4 - 3 \cr & = 4\,{\text{sq}}{\text{.}}\,{\text{units}} \cr} $$