A) \[|b|-|a|\]
B) \[|a|-|b|\]
C) \[|b|+|a|\]
D) \[-|b|-|a|\]
Correct Answer: A
Solution :
Case I. If\[0\le a<b\], then\[\frac{|x|}{x}=1\] \[\therefore \] \[I=\int_{a}^{b}{1\cdot dx=b-a=|b|-|a|}\] Case II: If\[a<b\le 0\], then\[|x|\,\,=-x\] \[\therefore \] \[I=\int_{a}^{b}{\frac{-x}{x}dx=\int_{a}^{b}{(-1)dx=[-x]_{a}^{b}}}\] \[=-b-(-a)\,=|b|\,-|a|\] Case lll lf a < 0 < b. then\[|x|\,\,=-x\] when\[a<x<0\] and\[|x|\,\,=x\] when\[0<x<b\] \[I=\int_{a}^{b}{\frac{|x|}{x}dx}\] \[=\int_{a}^{0}{\frac{|x|}{x}dx+\frac{|x|}{x}dx}\] \[=\int_{a}^{0}{\frac{-x}{x}dx+\int_{0}^{b}{\cdot \frac{x}{x}dx}}\] \[=\int_{a}^{0}{(-1)dx+\int_{0}^{b}{(1)dx}}\] \[=[-x]_{a}^{0}+[x]_{0}^{b}\] \[=a+b=b-(-a)\] \[=|b|-|a|\] Hence, in all cases, \[I=\int_{a}^{b}{\frac{|x|}{x}dx=|b|-|a|}\]
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