Question

Find the integral $\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $?

Answer 1

Hint: Before solving the question regenerate the limits because $\left| x \right|$ differs when $x \ge 0$ and $x < 0$.

Complete step by step solution:
$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $
Splitting the Mod.
$\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{x,x \ge 0}\\{ - x,x < 0}\end{array}} \right.$
Now For $\left| {x - 1} \right|$ ,
$\left| {x - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{\left( {x - 1} \right),x - 1 \ge 0}\\{ - \left( {x - 1} \right),x - 1 < 0}\\{1 - x,x < 1}\end{array}} \right.$
By using the above breaking of the Mod we will be changing the Limits for the integration,
$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $
$ = \int\limits_0^1 {\left| {1 - x - x} \right|dx + \int\limits_1^2 {\left| {x - 1 - x} \right|dx} } $
$ = \int\limits_0^1 {\left| {1 - 2x} \right|dx + \int\limits_1^2 {1.dx} } $
As, $\left| {1 - 2x} \right|$ is again a modulus function , Hence we will be breaking the mod for $\left| {1 - 2x} \right|$.
Which will be –
$\left| {1 - 2x} \right| = \left\{ {\begin{array}{*{20}{c}}{\left( {1 - 2x} \right),1 - 2x \ge 0}\\{ - \left( {1 - 2x} \right),x > \dfrac{1}{2}}\\{2x - 1}\end{array}} \right.$
Solving Further after breaking the mod $\left| {1 - 2x} \right|$
$\int\limits_0^{\frac{1}{2}} {1 - 2xdx + \int\limits_{\dfrac{1}{2}}^1 {2x - 1dx} } + {\left[ x \right]_1}^2$
$ = {\left[ x \right]_0}^{\frac{1}{2}} - 2{\left[ {\dfrac{{{x^2}}}{2}} \right]_0}^{\frac{1}{2}} + 2{\left[ {\dfrac{{{x^2}}}{2}} \right]_{\dfrac{1}{2}}}^1 - {\left[ x \right]_{\dfrac{1}{2}}}^1 + {\left[ x \right]^2}_1$
Now , Putting the limits.
$ = \dfrac{1}{2} - \dfrac{1}{4} + 1 - \dfrac{1}{4} - \dfrac{1}{2} + 1$
$ = 2 - \dfrac{1}{2}$
$ = \dfrac{3}{2}$

Hence the answer is $\dfrac{3}{2}$

Note: For such types of problems always split the modulus function and be very careful while changing the limits. If applying or changing limits is problematic for you, you can solve the integral up until the point when you need to replace the variable back in and use the existing limits.