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Question Number 118395 by TANMAY PANACEA last updated on 17/Oct/20

∫_a ^b ((f(x))/(f(x)+f(a+b−x)))dx

$$\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

((b−a)/2)

$$\frac{{b}−{a}}{\mathrm{2}} \\ $$

Commented by TANMAY PANACEA last updated on 17/Oct/20

thank you...durga pujo hochhe

$${thank}\:{you}...{durga}\:{pujo}\:{hochhe} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

From 23 oct

$${From}\:\mathrm{23}\:{oct} \\ $$

Commented by Dwaipayan Shikari last updated on 17/Oct/20

Are you in Kolkata sir?

$${Are}\:{you}\:{in}\:{Kolkata}\:{sir}? \\ $$

Commented by TANMAY PANACEA last updated on 17/Oct/20

na ami nagpur a..

$${na}\:{ami}\:{nagpur}\:{a}..\: \\ $$

Answered by mathmax by abdo last updated on 17/Oct/20

I =∫_a ^b ((f(x))/(f(x)+f(a+b−x)))dx we do the changement a+b−x=t ⇒ I =∫_b ^a ((f(a+b−t))/(f(a+b−t)+f(t)))(−dt) =∫_a ^b ((f(a+b−x))/(f(a+b−x)+f(x)))dx ⇒ 2I =∫_a ^b ((f(x)+f(a+b−x))/(f(a+b−x)+f(x)))dx =∫_a ^b dx =b−a ⇒I=((b−a)/2)

$$\mathrm{I}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \:\:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{a}+\mathrm{b}−\mathrm{x}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\mathrm{b}} ^{\mathrm{a}} \:\frac{\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{t}\right)}{\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)}\left(−\mathrm{dt}\right)\:=\int_{\mathrm{a}} ^{\mathrm{b}} \frac{\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{2I}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \:\frac{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{dx}\:=\mathrm{b}−\mathrm{a}\:\Rightarrow\mathrm{I}=\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}} \\ $$

Commented by TANMAY PANACEA last updated on 17/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Bird last updated on 17/Oct/20

you still work at railways sir tanmay..

$${you}\:{still}\:{work}\:{at}\:{railways}\:{sir}\:{tanmay}.. \\ $$

Commented by TANMAY PANACEA last updated on 17/Oct/20

yes sir till become senior citizen in year 2030

$${yes}\:{sir}\:{till}\:{become}\:{senior}\:{citizen}\:{in}\:{year}\:\mathrm{2030} \\ $$

Commented by mathmax by abdo last updated on 17/Oct/20

good luck sir tanmay

$$\mathrm{good}\:\mathrm{luck}\:\mathrm{sir}\:\mathrm{tanmay} \\ $$

Answered by TANMAY PANACEA last updated on 17/Oct/20

formula ∫_q ^p f(x)dx=∫_q ^p f(p+q−x)dx I=∫_a ^b ((f(x))/(f(x)+f(a+b−x)))dx =∫_a ^b ((f(a+b−x))/(f(a+b−x)+f{(a+b−(a+b−x)}))dx =∫_a ^b ((f(a+b−x))/(f(a+b−x)+f(x)))dx 2I=∫_a ^b dx I=((b−a)/2)

$${formula} \\ $$$$\int_{{q}} ^{{p}} {f}\left({x}\right){dx}=\int_{{q}} ^{{p}} {f}\left({p}+{q}−{x}\right){dx} \\ $$$${I}=\int_{{a}} ^{{b}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left({a}+{b}−{x}\right)}{dx} \\ $$$$=\int_{{a}} ^{{b}} \frac{{f}\left({a}+{b}−{x}\right)}{{f}\left({a}+{b}−{x}\right)+{f}\left\{\left({a}+{b}−\left({a}+{b}−{x}\right)\right\}\right.}{dx} \\ $$$$=\int_{{a}} ^{{b}} \frac{{f}\left({a}+{b}−{x}\right)}{{f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{{a}} ^{{b}} {dx} \\ $$$${I}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by bramlexs22 last updated on 17/Oct/20

this is formula King Integral sir?

$${this}\:{is}\:{formula}\:{King}\:{Integral}\:{sir}? \\ $$

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