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

∫_a ^b ((∣x∣)/x)dx

∫ab∣x∣xdx

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.

case1b⩾a⩾0⇒∫ab∣x∣xdx=∫abdx=b−acase2a⩽0andb⩾0⇒I=∫a0−dx+∫0bdx=a+bcase3a⩽b⩽0⇒I=∫ab−dx=a−bcase4a⩾0andb⩽0⇒I=∫a0∣x∣xdx+∫0b∣x∣xdx=−∫0adx−∫0b−dx=−a+b=b−a.

Answered by ajfour last updated on 12/Nov/18

=((x^2 /(∣x∣)))∣_a ^b =∣b∣−∣a∣.

=(x2∣x∣)∣ab=∣b∣−∣a∣.

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

excellent...

excellent...

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