Evaluate the given limit - \[\int\limits_{a}^{b}{\ s gn xdx}=\](where \[a,b\in R\])
A. \[\left| b \right|-\left| a \right|\]
B. \[\left( b-a \right)\ s gn \left( b-a \right)\]
C. \[b\ s gn b-a\ s gn a\]
D. \[\left| a \right|-\left| b \right|\]

Hint: Divide the question in the number of cases related to a and b with respect to signs of them. Then generalize the solution at last.

Complete step-by-step answer:
 Here, we have given expression/function as –
\[\int\limits_{a}^{b}{\ s gn xdx}=?\]
As we can define \[\ s gn x\]function as follows:
\[\ s gn x=\left\{ \begin{matrix}
   \dfrac{|x|}{x},x\ne 0 \\
   0,x=0 \\
\end{matrix} \right.\]
                 Or
\[\ s gn x=\left\{ \begin{matrix}
   1,x>0 \\
   -1,x<0 \\
   0,x=0 \\
\end{matrix} \right.\]
Representation on graph –


Case1: \[a\ge 0,b\ge 0\] \[a,b\in R\]
\[\int\limits_{a}^{b}{\ s gn x dx=\int\limits_{a}^{b}{1dx=b-a}}\]
As we can see from the graph or defined function that is a, b both are greater than 0 then \[\ s gn x=1\].

Case: 2 \[a\le 0,b\le 0\]
\[\int\limits_{a}^{b}{\ s gn (x)dx=\int\limits_{a}^{b}{-1dx=-(b-a)=a-b}}\]
Similarly, if \[a\le 0\And b\le 0\]then, \[\ s gn (x)=-1\].

Case: 3 \[a\ge 0,b\le 0\]
\[\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{\ s gn (x)dx}}\]

We have a property of integration as
\[\int\limits_{a}^{b}{f(x)dx=-\int\limits_{b}^{a}{f(x)dx}}\]
Hence, \[\int\limits_{a}^{b}{\ s gn (x)dx=-\int\limits_{b}^{a}{\ s gn (x)dx}=-\left[ \int\limits_{b}^{0}{\ s gn (x)dx}+\int\limits_{0}^{a}{\ s gn (x)dx} \right]}\]
Property used: - \[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{c}{f(x)dx}+\int\limits_{c}^{b}{f(x)dx}\] if \[aTherefore, expression will become - \[-\int\limits_{b}^{0}{\ s gn xdx}-\int\limits_{0}^{a}{\ s gn xdx}\]
As b is negative, so from b to 0 \[\ s gn (x)=-1\]and a is positive hence, \[\ s gn x=1\].
\[-\int\limits_{b}^{a}{(-1)dx}-\int\limits_{0}^{a}{1dx}=-b-a\]
Case: 4 \[a\le 0,b\ge 0\]

Similarly, by using integrating properties explained in case 3 we can proceed here as well: -
\[\begin{align}
  & \int\limits_{a}^{b}{\ s gn xdx}=\int\limits_{a}^{0}{\ s gn xdx}+\int\limits_{0}^{b}{\ s gn xdx} \\
 & =\int\limits_{a}^{0}{-1dx}+\int\limits_{0}^{b}{+1dx} \\
 & \Rightarrow a+b \\
\end{align}\]
Here, \[\int\limits_{a}^{b}{\ s gn xdx}=\left\{ \begin{matrix}
   b-a,a\ge 0,b\ge 0 \\
   a-b,a\le 0,b\le 0 \\
   -b-a,a\ge 0,b\le 0 \\
   a+b,a\le 0,b\ge 0 \\
\end{matrix} \right.\]
Now, we can generalise the solution by observation as \[|b|-|a|\And \left( b\ s gn b-a\ s gn a \right)\]from options.

\[|b|-|a|=\left\{ \begin{matrix}
   b-a,a\ge 0,b\ge 0 \\
   a-b,a\le 0,b\le 0 \\
   -b-a,a\ge 0,b\le 0 \\
   a+b,a\le 0,b\ge 0 \\
\end{matrix} \right.\]
We can note down that generalization is done on the basis of modulus function as follows: -
\[\begin{align}
  & |x|=\left\{ \begin{matrix}
   x,x>0 \\
   -x,x<0 \\
   0,x=0 \\
\end{matrix} \right. \\
 & b\ s gn b-a\ s gn a=\left\{ \begin{matrix}
   b-a,a\ge 0,b\ge 0 \\
   a-b,a\le 0,b\le 0 \\
   -b-a,a\ge 0,b\le 0 \\
   a+b,a\le 0,b\ge 0 \\
\end{matrix} \right. \\
\end{align}\]
Hence, option (a) and option (c) are correct.

Note: One can increase the number of cases by taking cases as a>0, b>0 and a>b or b