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Question Number 47607 by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

∫_a ^b ((∣x∣)/x)dx

$$\int_{{a}} ^{{b}} \frac{\mid{x}\mid}{{x}}{dx} \\ $$

Commented by maxmathsup by imad last updated on 12/Nov/18

case 1   b≥a≥0 ⇒ ∫_a ^b  ((∣x∣)/x)dx=∫_a ^b dx=b−a  case2 a≤0 and b≥0 ⇒I =∫_a ^0 −dx +∫_0 ^b dx =a+b  case 3  a≤b≤0 ⇒I =∫_a ^b −dx =a−b  case4 a≥0 and b≤0 ⇒ I =∫_a ^0  ((∣x∣)/x)dx +∫_0 ^b  ((∣x∣)/x)dx =−∫_0 ^a dx−∫_0 ^b  −dx  =−a+b =b−a.

$${case}\:\mathrm{1}\:\:\:{b}\geqslant{a}\geqslant\mathrm{0}\:\Rightarrow\:\int_{{a}} ^{{b}} \:\frac{\mid{x}\mid}{{x}}{dx}=\int_{{a}} ^{{b}} {dx}={b}−{a} \\ $$$${case}\mathrm{2}\:{a}\leqslant\mathrm{0}\:{and}\:{b}\geqslant\mathrm{0}\:\Rightarrow{I}\:=\int_{{a}} ^{\mathrm{0}} −{dx}\:+\int_{\mathrm{0}} ^{{b}} {dx}\:={a}+{b} \\ $$$${case}\:\mathrm{3}\:\:{a}\leqslant{b}\leqslant\mathrm{0}\:\Rightarrow{I}\:=\int_{{a}} ^{{b}} −{dx}\:={a}−{b} \\ $$$${case}\mathrm{4}\:{a}\geqslant\mathrm{0}\:{and}\:{b}\leqslant\mathrm{0}\:\Rightarrow\:{I}\:=\int_{{a}} ^{\mathrm{0}} \:\frac{\mid{x}\mid}{{x}}{dx}\:+\int_{\mathrm{0}} ^{{b}} \:\frac{\mid{x}\mid}{{x}}{dx}\:=−\int_{\mathrm{0}} ^{{a}} {dx}−\int_{\mathrm{0}} ^{{b}} \:−{dx} \\ $$$$=−{a}+{b}\:={b}−{a}. \\ $$

Answered by ajfour last updated on 12/Nov/18

=((x^2 /(∣x∣)))∣_a ^b =∣b∣−∣a∣.

$$=\left(\frac{{x}^{\mathrm{2}} }{\mid{x}\mid}\right)\mid_{{a}} ^{{b}} =\mid{b}\mid−\mid{a}\mid. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

excellent...

$${excellent}... \\ $$

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