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I have the value of a function's integral and two points of this function. I'm searching for the function. Is it possible to find it? $$\int F dx= a$$ $$F(b)= F(c)= d$$ $a$, $b$, $c$ and $d$ are known. I'm searching for $F$

HDE 226868
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  • i'm a mechanical engineer and i need it for an engineering stuff –  Apr 11 '15 at 22:39
  • ehh, it's a simple question, I've thrown my answer up. –  Apr 11 '15 at 22:45
  • Out of curiosity, what is the engineering application here? – HDE 226868 Apr 11 '15 at 23:06
  • ok, I'm new here and don't know about the notation, thank you. –  Apr 11 '15 at 23:10
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    i want set an application that gives the different dimensions used for the conception of a conveyor for bulk transporting, i'm organizing my calculation on microsoft excel, and i want to modelize the section of the bulk on the belt of the conveyor, for that i need the function of the extern generatrice of the bulk, and all i have is the area (ie the integral) and the two points (according to the standard of the belt conveyor) –  Apr 11 '15 at 23:23

2 Answers2

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$$F = 0$$

Essentially what is being asked is:

$$\int0dx = a$$ Where $a$ is a constant. If $F > 0$ then there would be an $x$ in the solution somewhere.

  • yes i understand, i can't copy the boundaries here, we have boundaries in the integral, the integral is between two values, so F=!0. i'm searching for the interpolation of this function, i don't know if it's possible or no, all what i have is the value of its integral and 2 points. –  Apr 11 '15 at 22:57
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    Better give me the values then mate, can't help you otherwise. The solution would be to differentiate the solution of the integral, which I have done, assuming a is constant, if it isn't then I need more detail. –  Apr 11 '15 at 23:01
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Any solution is not unique. Given any solution $f$ to $$ \int_{y}^{z} F(x) \, dx = a, \\ F(b)=F(c)=d, $$ the function $$ f(x) + x' (x'^2-b'^2)(x'^2-c'^2), $$ where $k'=k-(y+z)/2$, is also a solution (same values at $c,d$, and it is odd on the interval of integration, so has integral zero).

There is an easy way to find a solution, however. Let $f(x)=d+A(x-b)(x-c)$. Clearly this satisfies $f(b)=f(c)=d$. The integral is easy to give the correct value by choosing the right value of $A$: $$ a = \int_y^z f(x) \, dx = d(y-z)+A\left( \frac{z^3-y^3}{3}-(b+c)\frac{z^2-y^2}{2}+bc(z-y) \right) \\ = (y-z)\left( d+A\left(bc -(b+c) \frac{y+z}{2} + \frac{z^2+zy+y^2}{3} \right)\right) $$ If that's zero, try multiplying by f another power of $x$.

Chappers
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