Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 97007 by bagjamath last updated on 06/Jun/20

Commented by bagjamath last updated on 06/Jun/20

What is the theorm Sir?  ∫_0 ^1 y dx=∫_∞ ^0 x dy ??

$${What}\:{is}\:{the}\:{theorm}\:{Sir}? \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {y}\:{dx}=\int_{\infty} ^{\mathrm{0}} {x}\:{dy}\:?? \\ $$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

Itdoes not make any sense.

$$\mathrm{Itdoes}\:\mathrm{not}\:\mathrm{make}\:\mathrm{any}\:\mathrm{sense}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

Instead it can be written as  ∫_0 ^1 ydx=∫_0 ^1 (√(−1+(√((4/x)−3)))) dx

$$\mathrm{Instead}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ydx}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}}\:\mathrm{dx} \\ $$

Commented by bagjamath last updated on 06/Jun/20

Commented by bagjamath last updated on 06/Jun/20

This is question

$${This}\:{is}\:{question} \\ $$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

if x=(4/(3+(1+y^2 )^2 ))  then  dx = ((−16(y^3 +y))/((y^4 +2y^2 +4)^2 )) dy  then the integration turns to  ∫_0 ^∞ y.((16(y^3 +y))/((y^4 +2y^2 +4)^2 )) dy

$$\mathrm{if}\:\mathrm{x}=\frac{\mathrm{4}}{\mathrm{3}+\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\mathrm{then} \\ $$$$\mathrm{dx}\:=\:\frac{−\mathrm{16}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{y}\right)}{\left(\mathrm{y}^{\mathrm{4}} +\mathrm{2y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{dy} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{turns}\:\mathrm{to} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{y}.\frac{\mathrm{16}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{y}\right)}{\left(\mathrm{y}^{\mathrm{4}} +\mathrm{2y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{dy} \\ $$

Commented by mr W last updated on 06/Jun/20

it′s not a theorem, but only a technique.  technique see Q76146

$${it}'{s}\:{not}\:{a}\:{theorem},\:{but}\:{only}\:{a}\:{technique}. \\ $$$${technique}\:{see}\:{Q}\mathrm{76146} \\ $$

Commented by mr W last updated on 06/Jun/20

see also Q76680

$${see}\:{also}\:{Q}\mathrm{76680} \\ $$

Commented by mr W last updated on 06/Jun/20

Commented by mr W last updated on 06/Jun/20

the technique is following:  ∫_0 ^1 ydx=∫_0 ^1 f(x)dx is the area under  the curve (shaded area).  this area can also be calculated as  ∫_0 ^∞ xdy=∫_0 ^∞ f^(−1) (x)dx.  in our case the calculation of  ∫_0 ^∞ f^(−1) (x)dx is much easier than  ∫_0 ^1 f(x)dx.

$${the}\:{technique}\:{is}\:{following}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ydx}=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:{is}\:{the}\:{area}\:{under} \\ $$$${the}\:{curve}\:\left({shaded}\:{area}\right). \\ $$$${this}\:{area}\:{can}\:{also}\:{be}\:{calculated}\:{as} \\ $$$$\int_{\mathrm{0}} ^{\infty} {xdy}=\int_{\mathrm{0}} ^{\infty} {f}^{−\mathrm{1}} \left({x}\right){dx}. \\ $$$${in}\:{our}\:{case}\:{the}\:{calculation}\:{of} \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}^{−\mathrm{1}} \left({x}\right){dx}\:{is}\:{much}\:{easier}\:{than} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

yes sir it is the concept of inverse function.  Thank you. But I think in the question it should  be mentioned somewhere as x is a function of  y. Then this misunderstood can be avoided.

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{concept}\:\mathrm{of}\:\mathrm{inverse}\:\mathrm{function}. \\ $$$$\mathrm{Thank}\:\mathrm{you}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{think}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{it}\:\mathrm{should} \\ $$$$\mathrm{be}\:\mathrm{mentioned}\:\mathrm{somewhere}\:\mathrm{as}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of} \\ $$$$\mathrm{y}.\:\mathrm{Then}\:\mathrm{this}\:\mathrm{misunderstood}\:\mathrm{can}\:\mathrm{be}\:\mathrm{avoided}. \\ $$

Commented by bagjamath last updated on 06/Jun/20

Thank You Sir, Now i understand

$${Thank}\:{You}\:{Sir},\:{Now}\:{i}\:{understand} \\ $$

Commented by bagjamath last updated on 06/Jun/20

What application do you use, Sir?

$${What}\:{application}\:{do}\:{you}\:{use},\:{Sir}? \\ $$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

you can use GeoGebra or Desmos

$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{GeoGebra}\:\mathrm{or}\:\mathrm{Desmos} \\ $$

Commented by abdomathmax last updated on 07/Jun/20

I =∫_0 ^1 (√(−1+(√((4/x)−3))))dx we do the changement  (√(−1+(√((4/x)−3))))=t ⇒(√((4/x)−3))=t^2  +1 ⇒  (4/x) −3 =(t^2  +1)^(2 )  ⇒(4/x) =(t^(2 ) +1)^2 +3 ⇒  (x/4)=(1/((t^2  +1)^2  +3)) ⇒x =(4/((t^2  +1)^2  +3)) ⇒  dx =−4 ×((2(2t)(t^2 +1))/((t^2  +1)^(2 )  +3)) =−16 ×((t^3  +t)/((t^2  +1)^2  +3)) ⇒  I  =16∫_0 ^∞  t× ((t^3  +t)/((t^2  +1)^2  +3))dt  =16 ∫_0 ^∞   ((t^4  +t^2 )/((t^2  +1)^2  +3))dt  =8 ∫_(−∞) ^(+∞)  ((t^4  +t^2 )/((t^2  +1)^(2 ) +3))dt let ϕ(z) =((z^4  +z^2 )/((z^2  +1)^2 +3))  poles of ϕ?  ϕ(z) =((z^4  +z^2 )/((z^2 +1−i(√3))(z^2 +1+i(√3))))  −1+i(√3)=2(−(1/2)+i((√3)/2)) =2e^((2iπ)/3)  and− 1−i(√3)=2e^(−((2iπ)/3))   ϕ(z) =((z^4  +z^2 )/((z^2  −2e^((2iπ)/3) )(z^2 −2e^(−((i2π)/3)) )))  =((z^4  +z^2 )/((z−(√2)e^((iπ)/3) )(z+(√2)e^((iπ)/3) )(z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) )))  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ{ Res(ϕ,(√2)e^((iπ)/3) )+Res(ϕ,−(√2)e^((−iπ)/3) )}  Res(ϕ,(√2)e^((iπ)/3) ) =((4 e^((4iπ)/3)  +2 e^((i2π)/3) )/(4(√2)e^((iπ)/3) (2i sin(((2π)/3)))))  =((2 e^((4iπ)/3) +e^((i2π)/3) )/(4(√2)(i((√3)/2))))×e^(−((iπ)/3))    =((2 e^(iπ)  +e^((iπ)/3) )/(2i(√2)(√3))) =((−2+e^((iπ)/3)  )/(2i(√6)))  Res(ϕ,−(√2)e^((−iπ)/3) ) = ((4 e^((−4iπ)/(3 ))  +2 e^(−((2iπ)/3)) )/(−2(√2)e^(−((iπ)/3)) (2e^(−((2iπ)/3)) −2e^((2iπ)/3) )))  =((4 e^(−((4iπ)/3))  +2 e^(−((2iπ)/3)) )/(4(√2)(2isin(((2π)/3)))))×e^((iπ)/3)  = ((2 e^(−iπ)  +e^(−((iπ)/3)) )/(2(√3)(2i×((√3)/2))))  =((−2 +e^(−((iπ)/3)) )/(2i(√6)))  ....be continued....

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}}=\mathrm{t}\:\Rightarrow\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}=\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{4}}{\mathrm{x}}\:−\mathrm{3}\:=\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}\:} \:\Rightarrow\frac{\mathrm{4}}{\mathrm{x}}\:=\left(\mathrm{t}^{\mathrm{2}\:} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\:\Rightarrow \\ $$$$\frac{\mathrm{x}}{\mathrm{4}}=\frac{\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{4}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{dx}\:=−\mathrm{4}\:×\frac{\mathrm{2}\left(\mathrm{2t}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}\:} \:+\mathrm{3}}\:=−\mathrm{16}\:×\frac{\mathrm{t}^{\mathrm{3}} \:+\mathrm{t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{I}\:\:=\mathrm{16}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}×\:\frac{\mathrm{t}^{\mathrm{3}} \:+\mathrm{t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt} \\ $$$$=\mathrm{16}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{4}} \:+\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt} \\ $$$$=\mathrm{8}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{t}^{\mathrm{4}} \:+\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}\:} +\mathrm{3}}\mathrm{dt}\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}} \\ $$$$\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right)} \\ $$$$−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\mathrm{and}−\:\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} \:−\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}^{\mathrm{2}} −\mathrm{2e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)+\mathrm{Res}\left(\varphi,−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{4}\:\mathrm{e}^{\frac{\mathrm{4i}\pi}{\mathrm{3}}} \:+\mathrm{2}\:\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} }{\mathrm{4}\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2i}\:\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)} \\ $$$$=\frac{\mathrm{2}\:\mathrm{e}^{\frac{\mathrm{4i}\pi}{\mathrm{3}}} +\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} }{\mathrm{4}\sqrt{\mathrm{2}}\left(\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}×\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\:\:=\frac{\mathrm{2}\:\mathrm{e}^{\mathrm{i}\pi} \:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{2}}\sqrt{\mathrm{3}}}\:=\frac{−\mathrm{2}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:}{\mathrm{2i}\sqrt{\mathrm{6}}} \\ $$$$\mathrm{Res}\left(\varphi,−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\:\frac{\mathrm{4}\:\mathrm{e}^{\frac{−\mathrm{4i}\pi}{\mathrm{3}\:}} \:+\mathrm{2}\:\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{\mathrm{4}\:\mathrm{e}^{−\frac{\mathrm{4i}\pi}{\mathrm{3}}} \:+\mathrm{2}\:\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{4}\sqrt{\mathrm{2}}\left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}×\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:=\:\frac{\mathrm{2}\:\mathrm{e}^{−\mathrm{i}\pi} \:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{2i}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$=\frac{−\mathrm{2}\:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{6}}}\:\:....\mathrm{be}\:\mathrm{continued}.... \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 07/Jun/20

error in dx  i will delete the post!

$$\mathrm{error}\:\mathrm{in}\:\mathrm{dx}\:\:\mathrm{i}\:\mathrm{will}\:\mathrm{delete}\:\mathrm{the}\:\mathrm{post}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com