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Question Number 68019 by mathmax by abdo last updated on 03/Sep/19
let F(x) =∫_x ^(x^2 +1) e^(−2t) sin(xt)dt determine F^′ (x) and calculate lim_(x→0) F(x).
letF(x)=∫xx2+1e−2tsin(xt)dtdetermineF′(x)andcalculatelimx→0F(x).
Commented by kaivan.ahmadi last updated on 05/Sep/19
thank you sir abdo
thankyousirabdo
Commented by mathmax by abdo last updated on 05/Sep/19
sir ahmadi your answer is not correct because the function under integral contain 2 varibles x and t ...!
sirahmadiyouranswerisnotcorrectbecausethefunctionunderintegralcontain2variblesxandt…!
Commented by mathmax by abdo last updated on 05/Sep/19
F(x) =∫_x ^(x^2 +1) e^(−2t) sin(xt)dt we have u(x)=x and v(x)=x^2 +1and g(x,t) =e^(−2t) sin(xt) we use the formulae F^, (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v)−u^′ g(x,u) =∫_x ^(x^2 +1) t e^(−2t) cos(xt)dt +2x g(x,x^2 +1)−g(x,x^2 ) =∫_x ^(x^2 +1) t e^(−2t) cos(xt)dt +2xe^(−2(x^2 +1)) sin(x(x^2 +1))−e^(−2x^2 ) sin(x^3 ) we have ∫_x ^(x^2 +1 ) t e^(−2t) cos(xt)dt =Re(∫_x ^(x^2 +1) t e^(−2t+ixt) dt) ∫_x ^(x^2 +1) t e^((−2+ix)t) dt =[(t/(−2+ix)) e^((−2+ix)t) ]_(t=x) ^(t=x^2 +1) −∫_x ^(x^2 +1) (1/(−2+ix))e^((−2+ix)t) dt =(1/(−2+ix)){(x^2 +1)e^((−2+ix)(x^2 +1)) −xe^((−2+ix)x) } −(1/(−2+ix)) ∫_x ^(x^2 +1) e^((−2+ix)t) dt ....rest to finich the calculus.
F(x)=∫xx2+1e−2tsin(xt)dtwehaveu(x)=xandv(x)=x2+1andg(x,t)=e−2tsin(xt)weusetheformulaeF,(x)=∫u(x)v(x)∂g∂x(x,t)dt+v′g(x,v)−u′g(x,u)=∫xx2+1te−2tcos(xt)dt+2xg(x,x2+1)−g(x,x2)=∫xx2+1te−2tcos(xt)dt+2xe−2(x2+1)sin(x(x2+1))−e−2x2sin(x3)wehave∫xx2+1te−2tcos(xt)dt=Re(∫xx2+1te−2t+ixtdt)∫xx2+1te(−2+ix)tdt=[t−2+ixe(−2+ix)t]t=xt=x2+1−∫xx2+11−2+ixe(−2+ix)tdt=1−2+ix{(x2+1)e(−2+ix)(x2+1)−xe(−2+ix)x}−1−2+ix∫xx2+1e(−2+ix)tdt….resttofinichthecalculus.
Commented by mathmax by abdo last updated on 06/Sep/19
you are welcome sir.
youarewelcomesir.
Commented by mathmax by abdo last updated on 08/Sep/19
we have F(x)=∫_x ^(x^2 +1) e^(−2t) sin(xt)dt ⇒F(x)=Im(∫_x ^(x^2 +1) e^(−2t+ixt) dt) ∫_x ^(x^2 +1) e^((−2+ix)t) dt =[(1/(−2+ix)) e^((−2+ix)t) ]_x ^(x^2 +1) =((−1)/(2−ix)){ e^((−2+ix)(x^2 +1)) −e^((−2+ix)x) } =((−1)/(2−ix)){ e^(−2(x^2 +1)) e^(i(x^3 +x)) −e^(−2x) e^(ix^2 ) } =((−1)/(2−ix)){ e^(−2x^2 −2) (cos(x^3 +x)+isin(x^3 +x)−e^(−2x) (cos(x^2 )+isin(x^2 )} =−((2+ix)/(4+x^2 )){ e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 ) +i( e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )} =−(1/(4+x^2 ))(2+ix){......} =−(1/(4+x^2 )){ 2( e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 )) +2i(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )) +ix( e^(−2x^2 −2) cos(x^3 +x) −e^(−2x) cos(x^2 )−x(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )} ⇒ F(x) =−(1/(4+x^2 )){ 2(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )) +x(e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 )} ⇒lim_(x→0) F(x) =0
wehaveF(x)=∫xx2+1e−2tsin(xt)dt⇒F(x)=Im(∫xx2+1e−2t+ixtdt)∫xx2+1e(−2+ix)tdt=[1−2+ixe(−2+ix)t]xx2+1=−12−ix{e(−2+ix)(x2+1)−e(−2+ix)x}=−12−ix{e−2(x2+1)ei(x3+x)−e−2xeix2}=−12−ix{e−2x2−2(cos(x3+x)+isin(x3+x)−e−2x(cos(x2)+isin(x2)}=−2+ix4+x2{e−2x2−2cos(x3+x)−e−2xcos(x2)+i(e−2x2−2sin(x3+x)−e−2xsin(x2)}=−14+x2(2+ix){……}=−14+x2{2(e−2x2−2cos(x3+x)−e−2xcos(x2))+2i(e−2x2−2sin(x3+x)−e−2xsin(x2))+ix(e−2x2−2cos(x3+x)−e−2xcos(x2)−x(e−2x2−2sin(x3+x)−e−2xsin(x2)}⇒F(x)=−14+x2{2(e−2x2−2sin(x3+x)−e−2xsin(x2))+x(e−2x2−2cos(x3+x)−e−2xcos(x2)}⇒limx→0F(x)=0

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