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Question Number 67972 by mathmax by abdo last updated on 02/Sep/19
if F(x)=∫_(u(x)) ^(v(x)) g(x,t)dt determine a expression for F^′ (x).
ifF(x)=∫u(x)v(x)g(x,t)dtdetermineaexpressionforF′(x).
Answered by Tanmay chaudhury last updated on 03/Sep/19
(dF/dx)=∫_(u(x)) ^(v(x)) (∂/∂x)(g(x,t)dx)+((dv(x))/dx)g(x,v(x))−((du(x))/dx)g(x,u(x))
dFdx=∫u(x)v(x)∂∂x(g(x,t)dx)+dv(x)dxg(x,v(x))−du(x)dxg(x,u(x))
Commented by mathmax by abdo last updated on 03/Sep/19
can you sind the sourse of this formula sir tanmay or send the proof ...
canyousindthesourseofthisformulasirtanmayorsendtheproof…
Commented by Tanmay chaudhury last updated on 03/Sep/19
ok sir...
oksir…
Commented by mind is power last updated on 03/Sep/19
in my comment F(x)=∫_v ^u g(x,t)dt
inmycommentF(x)=∫vug(x,t)dt
Commented by mathmax by abdo last updated on 03/Sep/19
let verify this formulae by F(x) =∫_x ^x^2 e^(−xt) dt we have g(x,t) =e^(−xt) , u(x)=x and v(x)=x^2 gormulae give F^′ (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v)−u^′ g(x,u) =∫_x ^x^2 −t e^(−xt) dt +(2x)e^(−x^3 ) −e^(−x^2 ) by partswe get ∫_x ^x^2 t e^(−xt) dt =[−(t/x)e^(−xt) ]_x ^x^2 −∫_x ^x^2 (−(1/x))e^(−xt) dt =e^(−x^2 ) −x e^(−x^3 ) +(1/x) ∫_x ^x^2 e^(−xt) dt =e^(−x^2 ) −x e^(−x^3 ) +(1/x)[−(1/x)e^(−xt) ]_x ^x^2 =e^(−x^2 ) −xe^(−x^3 ) −(1/x^2 ){ e^(−x^3 ) −e^(−x^2 ) } =(1+(1/x^2 ))e^(−x^2 ) −(x+(1/x^2 ))e^(−x^3 ) ⇒F^′ (x) =−(1+(1/x^2 ))e^(−x^2 ) +(x+(1/x^2 ))e^(−x^3 ) +2x e^(−x^3 ) −e^(−x^2 ) =−(2+(1/x^2 ) )e^(−x^2 ) (3x+(1/x^2 ))e^(−x^3 ) now let find F(x) directly F(x) =[−(1/x)e^(−xt) ]_x ^x^2 =−(1/x){ e^(−x^3 ) −e^(−x^2 ) } =(1/x){e^(−x^2 ) −e^(−x^3 ) } ⇒ F^′ (x) =−(1/x^2 ){ e^(−x^2 ) −e^(−x^3 ) } +(1/x){−2x e^(−x^2 ) +3x^2 e^(−x^3 ) } =−(1/x^2 )e^(−x^2 ) +(1/x^2 )e^(−x^3 ) −2e^(−x^2 ) +3x e^(−x^3 ) =−((1/x^2 ) +2)e^(−x^2 ) +(3x+(1/x^2 ))e^(−x^2 ) we get the same result so the formulae is correct .
letverifythisformulaebyF(x)=∫xx2e−xtdtwehaveg(x,t)=e−xt,u(x)=xandv(x)=x2gormulaegiveF′(x)=∫u(x)v(x)∂g∂x(x,t)dt+v′g(x,v)−u′g(x,u)=∫xx2−te−xtdt+(2x)e−x3−e−x2bypartsweget∫xx2te−xtdt=[−txe−xt]xx2−∫xx2(−1x)e−xtdt=e−x2−xe−x3+1x∫xx2e−xtdt=e−x2−xe−x3+1x[−1xe−xt]xx2=e−x2−xe−x3−1x2{e−x3−e−x2}=(1+1x2)e−x2−(x+1x2)e−x3⇒F′(x)=−(1+1x2)e−x2+(x+1x2)e−x3+2xe−x3−e−x2=−(2+1x2)e−x2(3x+1x2)e−x3nowletfindF(x)directlyF(x)=[−1xe−xt]xx2=−1x{e−x3−e−x2}=1x{e−x2−e−x3}⇒F′(x)=−1x2{e−x2−e−x3}+1x{−2xe−x2+3x2e−x3}=−1x2e−x2+1x2e−x3−2e−x2+3xe−x3=−(1x2+2)e−x2+(3x+1x2)e−x2wegetthesameresultsotheformulaeiscorrect.
Answered by mind is power last updated on 03/Sep/19
let G(x,t)=∫g(x,t)dt F(x)=G(x,v(x))−G(x,u(x)) F′(x)=(∂/∂x){G(x,v(x))−G(x,u(x))}..1 (∂/∂x)(f(u,v))=(∂u/∂x).((∂f(u,v))/∂x)+(∂v/∂x).((∂f(u,v))/∂v) we apply this in 1 F′(x)= (∂/∂x)G(x,u(x))+(∂u/∂x) ((∂ G(x,u(x)))/∂t)−((∂G(x,v(x))/∂x)−(∂v/∂x).((∂G(x,v(x))/∂t) withe ((∂G(x,t))/∂t)=g(x,t) ((∂G(x,t))/∂x)=(∂/∂x)∫g(x,t)dt=∫((∂g(x,t))/∂x)dt we get ∫_0 ^u ((∂g(x,t)dt)/∂x)+u^′ g(x,u)−∫_0 ^v ((∂g(x,t))/∂x)dt−v^, g(x,v) =∫_v ^u ((∂g(x,t))/∂x)dt+u^′ g(x,u)−v^′ g(x,v)
letG(x,t)=∫g(x,t)dtF(x)=G(x,v(x))−G(x,u(x))F′(x)=∂∂x{G(x,v(x))−G(x,u(x))}..1∂∂x(f(u,v))=∂u∂x.∂f(u,v)∂x+∂v∂x.∂f(u,v)∂vweapplythisin1F′(x)=∂∂xG(x,u(x))+∂u∂x∂G(x,u(x))∂t−∂G(x,v(x)∂x−∂v∂x.∂G(x,v(x)∂twithe∂G(x,t)∂t=g(x,t)∂G(x,t)∂x=∂∂x∫g(x,t)dt=∫∂g(x,t)∂xdtweget∫0u∂g(x,t)dt∂x+u′g(x,u)−∫0v∂g(x,t)∂xdt−v,g(x,v)=∫vu∂g(x,t)∂xdt+u′g(x,u)−v′g(x,v)
Commented by mathmax by abdo last updated on 10/Sep/19
thank you sir.
thankyousir.

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