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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−2b) ^(2a−b) ∣(√((x+3b)((2a−b−x)))  −(√((3a−x)(2b−a+x))) ∣dx  ∫_(a−2b) ^(2a−b) (√((3a−3b−x+3b)×(2a−b−(3a−3b−x)))   −(√((3a−x)(2b−a+x))) ∣dx  ∫_(a−2b) ^(2a−b) ∣(√((3a−x)(2b−a−x))) −(√((x+3b)(2a−b−x))) dx  ∫_(a−2b) ^(2a−b) ∣p−q∣dx=∫_(a−2b) ^(2a−b) ∣q−p∣dx  that means p−q=q−p  2p=2q    p=q  (x+3b)(2a−b−x)=(3a−x)(2b−a−x)  2ax−bx−x^2 +6ab−3b^2 −3bx=6ab−3a^2 −3ax−2bx+ax+x^2   2ax+2ax−4bx+2bx−x^2 −x^2 +6ab−6ab=0  4ax−2bx−2x^2 =0  2x(2a−b−x)=0  x=2a−b

$$\int_{{a}−\mathrm{2}{b}} ^{\mathrm{2}{a}−{b}} \mid\sqrt{\left({x}+\mathrm{3}{b}\right)\left(\left(\mathrm{2}{a}−{b}−{x}\right)\right.}\:\:−\sqrt{\left(\mathrm{3}{a}−{x}\right)\left(\mathrm{2}{b}−{a}+{x}\right)}\:\mid{dx} \\ $$$$\int_{{a}−\mathrm{2}{b}} ^{\mathrm{2}{a}−{b}} \sqrt{\left(\mathrm{3}{a}−\mathrm{3}{b}−{x}+\mathrm{3}{b}\right)×\left(\mathrm{2}{a}−{b}−\left(\mathrm{3}{a}−\mathrm{3}{b}−{x}\right)\right.}\:\:\:−\sqrt{\left(\mathrm{3}{a}−{x}\right)\left(\mathrm{2}{b}−{a}+{x}\right)}\:\mid{dx} \\ $$$$\int_{{a}−\mathrm{2}{b}} ^{\mathrm{2}{a}−{b}} \mid\sqrt{\left(\mathrm{3}{a}−{x}\right)\left(\mathrm{2}{b}−{a}−{x}\right)}\:−\sqrt{\left({x}+\mathrm{3}{b}\right)\left(\mathrm{2}{a}−{b}−{x}\right)}\:{dx} \\ $$$$\int_{{a}−\mathrm{2}{b}} ^{\mathrm{2}{a}−{b}} \mid{p}−{q}\mid{dx}=\int_{{a}−\mathrm{2}{b}} ^{\mathrm{2}{a}−{b}} \mid{q}−{p}\mid{dx} \\ $$$${that}\:{means}\:{p}−{q}={q}−{p} \\ $$$$\mathrm{2}{p}=\mathrm{2}{q}\:\:\:\:{p}={q} \\ $$$$\left({x}+\mathrm{3}{b}\right)\left(\mathrm{2}{a}−{b}−{x}\right)=\left(\mathrm{3}{a}−{x}\right)\left(\mathrm{2}{b}−{a}−{x}\right) \\ $$$$\mathrm{2}{ax}−{bx}−{x}^{\mathrm{2}} +\mathrm{6}{ab}−\mathrm{3}{b}^{\mathrm{2}} −\mathrm{3}{bx}=\mathrm{6}{ab}−\mathrm{3}{a}^{\mathrm{2}} −\mathrm{3}{ax}−\mathrm{2}{bx}+{ax}+{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{ax}+\mathrm{2}{ax}−\mathrm{4}{bx}+\mathrm{2}{bx}−{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{6}{ab}−\mathrm{6}{ab}=\mathrm{0} \\ $$$$\mathrm{4}{ax}−\mathrm{2}{bx}−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{x}\left(\mathrm{2}{a}−{b}−{x}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}{a}−{b} \\ $$$$ \\ $$$$ \\ $$

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