Equation with definite integral.












1














Suppose that I have a function $f(x,y)$,



I want to solve



$$int_a^bf(x,y)dx =0.$$



w.r.t $y$.
Is it possible to say that this is equivalent to solve



$$f(b,y)-f(a,y) =0$$



since I can multiply on both the sides by $frac{d}{dx}$?






definite-integrals







share|cite|improve this question


















  • 1




    I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
    – Sam
    Nov 27 at 12:18








  • 1




    No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
    – GEdgar
    Nov 27 at 13:19










  • I see, thanks @GEdgar.
    – Sam
    Nov 27 at 14:04
















1














Suppose that I have a function $f(x,y)$,



I want to solve



$$int_a^bf(x,y)dx =0.$$



w.r.t $y$.
Is it possible to say that this is equivalent to solve



$$f(b,y)-f(a,y) =0$$



since I can multiply on both the sides by $frac{d}{dx}$?






definite-integrals







share|cite|improve this question


















  • 1




    I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
    – Sam
    Nov 27 at 12:18








  • 1




    No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
    – GEdgar
    Nov 27 at 13:19










  • I see, thanks @GEdgar.
    – Sam
    Nov 27 at 14:04














1












1








1







Suppose that I have a function $f(x,y)$,



I want to solve



$$int_a^bf(x,y)dx =0.$$



w.r.t $y$.
Is it possible to say that this is equivalent to solve



$$f(b,y)-f(a,y) =0$$



since I can multiply on both the sides by $frac{d}{dx}$?






definite-integrals







share|cite|improve this question













Suppose that I have a function $f(x,y)$,



I want to solve



$$int_a^bf(x,y)dx =0.$$



w.r.t $y$.
Is it possible to say that this is equivalent to solve



$$f(b,y)-f(a,y) =0$$



since I can multiply on both the sides by $frac{d}{dx}$?






definite-integrals




definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 at 11:50









Sam

869




869








  • 1




    I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
    – Sam
    Nov 27 at 12:18








  • 1




    No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
    – GEdgar
    Nov 27 at 13:19










  • I see, thanks @GEdgar.
    – Sam
    Nov 27 at 14:04














  • 1




    I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
    – Sam
    Nov 27 at 12:18








  • 1




    No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
    – GEdgar
    Nov 27 at 13:19










  • I see, thanks @GEdgar.
    – Sam
    Nov 27 at 14:04








1




1




I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
– Sam
Nov 27 at 12:18






I already have $f(x,y)$ and I want to find $y$ which satisfies the first equation. Since computing the integral might be impossible, I was wondering if the solution of the second equation (w.r.t $y$) is the same as the solution of the first equation.
– Sam
Nov 27 at 12:18






1




1




No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
– GEdgar
Nov 27 at 13:19




No, differentiation is no good for this. In general, when you solve $f(x)=0$ you get just one point, and it does not follow that $frac{d}{dx}f(x)=0$ at that one point. That is, the point where a graph crosses the $x$-axis is not also a point where the tangent to the graph is horizontal.
– GEdgar
Nov 27 at 13:19












I see, thanks @GEdgar.
– Sam
Nov 27 at 14:04




I see, thanks @GEdgar.
– Sam
Nov 27 at 14:04










1 Answer
1






active

oldest

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2














Your assertion is not correct.



Counterexample



Consider $f(x,y)=cos(x+y),$ $$int_{0}^{pi}cos(x+y)=sin(pi+y)-sin y=-2sin y.$$ For $y={pi};$ takes the integral the value $0.$
But
$$cos(pi+pi)-cos (0+pi)=2neq 0.$$






share|cite|improve this answer





















  • Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
    – Sam
    Nov 27 at 13:03








  • 1




    The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
    – user376343
    Nov 27 at 13:17













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Your assertion is not correct.



Counterexample



Consider $f(x,y)=cos(x+y),$ $$int_{0}^{pi}cos(x+y)=sin(pi+y)-sin y=-2sin y.$$ For $y={pi};$ takes the integral the value $0.$
But
$$cos(pi+pi)-cos (0+pi)=2neq 0.$$






share|cite|improve this answer





















  • Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
    – Sam
    Nov 27 at 13:03








  • 1




    The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
    – user376343
    Nov 27 at 13:17


















2














Your assertion is not correct.



Counterexample



Consider $f(x,y)=cos(x+y),$ $$int_{0}^{pi}cos(x+y)=sin(pi+y)-sin y=-2sin y.$$ For $y={pi};$ takes the integral the value $0.$
But
$$cos(pi+pi)-cos (0+pi)=2neq 0.$$






share|cite|improve this answer





















  • Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
    – Sam
    Nov 27 at 13:03








  • 1




    The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
    – user376343
    Nov 27 at 13:17
















2












2








2






Your assertion is not correct.



Counterexample



Consider $f(x,y)=cos(x+y),$ $$int_{0}^{pi}cos(x+y)=sin(pi+y)-sin y=-2sin y.$$ For $y={pi};$ takes the integral the value $0.$
But
$$cos(pi+pi)-cos (0+pi)=2neq 0.$$






share|cite|improve this answer












Your assertion is not correct.



Counterexample



Consider $f(x,y)=cos(x+y),$ $$int_{0}^{pi}cos(x+y)=sin(pi+y)-sin y=-2sin y.$$ For $y={pi};$ takes the integral the value $0.$
But
$$cos(pi+pi)-cos (0+pi)=2neq 0.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 12:37









user376343

2,8182822




2,8182822












  • Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
    – Sam
    Nov 27 at 13:03








  • 1




    The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
    – user376343
    Nov 27 at 13:17




















  • Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
    – Sam
    Nov 27 at 13:03








  • 1




    The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
    – user376343
    Nov 27 at 13:17


















Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
– Sam
Nov 27 at 13:03






Thanks! I suspected it. But can you explain to me why is not that? If my integral was indefinite I could have done $int f(x) dx = 0 implies int frac{d}{dx}f(x) dx = frac{d}{dx} 0 implies f(x) +c = 0$ but usually the $c$ cancel out when I have definite integral
– Sam
Nov 27 at 13:03






1




1




The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
– user376343
Nov 27 at 13:17






The result is a function of $y,$ say $G(y)=int_a^bf(x,y)dx.$ You can derivate it or integrate, but w.r.t. $y.$ If you post a new question with your function $f(x,y)$ it is not excluded that MSE solves the integral.
– user376343
Nov 27 at 13:17




















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