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Show that`int_a^b(|x|)/x dx=|b|=|a|dot` |
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Answer» Case I: If `0lealtb`, then `|x|//x=1` `:.I=int_(a)^(b)1dx=b-a=|b|-|a|` Case II: If `altble0`, then `|x|=-x` `:. I=int_(a)^(b)(-x)/xdx=int_(a)^(b)(-1)dx` `=[-x]_(a)^(b)=b-(-a)=|b|-|a|` Case III: If `alt0ltb` then `|x|=-x` when `altxlt0` and `|x|=x`, when `altxltb` `:.I=int_(a)^(|x|)/xdx=int_(a)^(0)(|x|)/x dx+int_(0)^(b)(|x|)/x dx` `=int_(a)^(0)(-x)/x dx+int_(0)^(b) 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 the cases `I=int_(a)^(b)(|x|)/x dx=|b|-|a|`. |
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