If \[f\left( x \right) = \left| {x - 1} \right|\], then find the value of \[\int_0^2 {f\left( x \right)dx} \].
A. 1
B. 0
C. 2
D. 12

Hint: First we will find the intervals where the value \[\left| {x - 1} \right|\] is positive and negative. Then rewrite the integration as the sum of integration. After that integrate it to get the value of \[\int_0^2 {f\left( x \right)dx} \].



Formula Used:Property of definite integration:
\[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \] where \[b < c < a\]



Complete step by step solution:We know that, \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{x < 0}\\{x,}&{x \ge 0}\end{array}} \right.\]
Case 1: When \[x - 1 < 0\]
\[x - 1 < 0\]
Add 1 on both sides:
\[ \Rightarrow x - 1 + 1 < 0 + 1\]
\[ \Rightarrow x < 1\]
Therefore, \[x - 1 < 0\] when \[x < 1\]
Case 2: When \[x - 1 > 0\]
\[x - 1 > 0\]
Add 1 on both sides:
\[ \Rightarrow x - 1 + 1 > 0 + 1\]
\[ \Rightarrow x > 1\]
Therefore, \[x - 1 > 0\] when \[x > 1\]
The value of \[\left| {x - 1} \right|\] is positive when \[x > 1\] and the value of \[\left| {x - 1} \right|\] is negative when \[x < 1\].
In other words \[\left| {x - 1} \right|\] is positive when \[1 < x < 2\]and \[\left| {x - 1} \right|\] is negative when \[0 < x < 1\].
Now we will rewrite the given integration as the sum of two integrations. The limits of the first integration will be 0 to 1 and the limits of the second integration will be 1 to 2.
\[\int_0^2 {f\left( x \right)dx} \]
\[ = \int_0^1 {f\left( x \right)dx} + \int_1^2 {f\left( x \right)dx} \]
\[ = \int_0^1 {\left| {x - 1} \right|dx} + \int_1^2 {\left| {x - 1} \right|dx} \]
We know that \[\left| {x - 1} \right|\] is positive when \[1 < x < 2\]and \[\left| {x - 1} \right|\] is negative when \[0 < x < 1\].
Thus rewrite the above expression:
\[ = \int_0^1 { - \left( {x - 1} \right)dx} + \int_1^2 {\left( {x - 1} \right)dx} \]
\[ = - \left[ {\dfrac{{{x^2}}}{2} - x} \right]_0^1 + \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^2\]
\[ = - \left[ {\dfrac{{{1^2}}}{2} - 1 + \left( {\dfrac{{{0^2}}}{2} - 0} \right)} \right] + \left[ {\dfrac{{{2^2}}}{2} - 2 - \left( {\dfrac{{{1^2}}}{2} - 1} \right)} \right]\]
\[ = - \dfrac{1}{2} + 1 + \left[ {2 - 2 - \left( {\dfrac{1}{2} - 1} \right)} \right]\]
\[ = - \dfrac{1}{2} + 1 + \dfrac{1}{2}\]
\[ = 1\]



Option ‘A’ is correct



Note: To solve the integration of absolute function, first we need to find the positive and negative values in the given limits. Students often skip this concept. They start to solve it directly means without considering the positive and negative values of the function. This is incorrect way.