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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 $\frac{d}{dx}$?

Sam
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    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 '18 at 12:18
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    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 '18 at 13:19
  • I see, thanks @GEdgar. – Sam Nov 27 '18 at 14:04

1 Answers1

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Your assertion is not correct.

Counterexample

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

user376343
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  • 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 '18 at 13:03
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    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 '18 at 13:17