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Question Number 45495 by Meritguide1234 last updated on 13/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
∫a−2b2a−b∣(x+3b)((2a−b−x)−(3a−x)(2b−a+x)∣dx∫a−2b2a−b(3a−3b−x+3b)×(2a−b−(3a−3b−x)−(3a−x)(2b−a+x)∣dx∫a−2b2a−b∣(3a−x)(2b−a−x)−(x+3b)(2a−b−x)dx∫a−2b2a−b∣p−q∣dx=∫a−2b2a−b∣q−p∣dxthatmeansp−q=q−p2p=2qp=q(x+3b)(2a−b−x)=(3a−x)(2b−a−x)2ax−bx−x2+6ab−3b2−3bx=6ab−3a2−3ax−2bx+ax+x22ax+2ax−4bx+2bx−x2−x2+6ab−6ab=04ax−2bx−2x2=02x(2a−b−x)=0x=2a−b
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