Question

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. $\left| a \right|-\left| b \right|$

Answer 1

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:

\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{aligned} & \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{aligned}$$ Here, $$\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{aligned} & |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{aligned}$$ _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