Question
By using the properties of definite integrals, evaluate the integral $\int\limits_0^4 {\left| {x - 1} \right|dx} $

Answer

Let $I = \int\limits_0^4 {\left| {x - 1} \right|dx} $ ...(i)
Here x - 1 = 0
$ \Rightarrow x = 1 \in \left( {0,4} \right)$
$\therefore$ From eq. (i),
$I = \int\limits_0^1 {\left| {x - 1} \right|dx + \int\limits_1^4 {\left| {x - 1} \right|} dx} $
$= - \int\limits_0^1 {\left( {x - 1} \right)dx + \int\limits_1^4 {\left( {x - 1} \right)} dx} $
$ \Rightarrow I = - \left( {\frac{{{x^2}}}{2} - x} \right)_0^1 + \left( {\frac{{{x^2}}}{2} - x} \right)_1^4$
$= - \left\{ {\left( {\frac{1}{2} - 1} \right) - 0} \right\} + \left\{ {\frac{{16}}{2} - 4 - \left( {\frac{1}{2} - 1} \right)} \right\}$

$= \frac{1}{2} + 8 - 4 + \frac{1}{2}$
= 9 - 4
= 5

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