What is the value of the definite integral \[\int\limits_0^2 {\left| {x - 1} \right|} dx\]?
A. 0
B. 2
C. \[\dfrac{1}{2}\]
D. 1

Hint: Here, a definite integral is given. The function present in the integral is an absolute value function. First, apply the absolute function formula that is \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]. Then, break the interval for the absolute value function. After that, solve the integrals. In the end, apply the limits and solve it to get the required answer.



Formula Used:Absolute value function: \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_0^2 {\left| {x - 1} \right|} dx\].
Let consider,
\[I = \int\limits_0^2 {\left| {x - 1} \right|} dx\]
Substitute \[x = x - 1\] in the absolute value function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{if x < 0}\\{x,}&{if x \ge 0}\end{array}} \right.\].
We get,
\[\left| {x - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - 1} \right),}&{if x - 1 < 0}\\{x - 1,}&{if x - 1 \ge 0}\end{array}} \right.\]
\[ \Rightarrow \left| {x - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - 1} \right),}&{if x < 1}\\{x - 1,}&{if x \ge 1}\end{array}} \right.\]
The interval of the integration is 0 to 2.
This implies that \[\left| {x - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{ - \left( {x - 1} \right),}&{if 0 < x < 1}\\{x - 1,}&{if 1 \le x < 2}\end{array}} \right.\].
Break the integration by using the above absolute value function.
\[I = \int\limits_0^1 { - \left( {x - 1} \right)} dx + \int\limits_1^2 {\left( {x - 1} \right)} dx\]
\[ \Rightarrow I = \int\limits_0^1 {\left( {1 - x} \right)} dx + \int\limits_1^2 {\left( {x - 1} \right)} dx\]
Now solve the integrals by using the formulas \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\] and \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[ \Rightarrow I = \left[ {x - \dfrac{{{x^2}}}{2}} \right]_0^1 + \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^2\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\left( {1 - \dfrac{{{1^2}}}{2}} \right) - \left( {0 - \dfrac{{{0^2}}}{2}} \right)} \right] + \left[ {\left( {\dfrac{{{2^2}}}{2} - 2} \right) - \left( {\dfrac{{{1^2}}}{2} - 1} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{2} + \left[ {\left( {2 - 2} \right) - \left( { - \dfrac{1}{2}} \right)} \right]\]
\[ \Rightarrow I = \dfrac{1}{2} + \left[ {0 + \dfrac{1}{2}} \right]\]
\[ \Rightarrow I = \dfrac{1}{2} + \dfrac{1}{2}\]
\[ \Rightarrow I = 1\]
Thus, \[\int\limits_0^2 {\left| {x - 1} \right|} dx = 1\].



Option ‘D’ is correct


Note: Students often do mistake to integrating \[\int\limits_a^b {{x^n}dx} \]. They apply the formula \[\int\limits_a^b {{x^n}dx = \left[ {{x^{n + 1}}} \right]} _a^b\] which is an incorrect formula. The correct formula is \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].