Integration equals another integration $\displaystyle\int f(x)\ dx=\displaystyle\int g(x)\ dx$
So what is the relation between $f$ and $g$ 
I found this solution but i am not sure it is right or not :
$\displaystyle\int (f(x)-g(x))\ dx=0$.
So $f(x)-g(x)= C$
 A: If you put $h(x) = f(x) - g(x)$ (and $f(x), g(x)$ are sufficiently smooth functions) then if $\frac{dh(x)}{dx}=0$ then you can say $h(x)$ is a constant (not necessarily zero) independent of $x$. However, if you have $\int_a^b h(x) \, dx = 0$ for all limits $a, b$ then you can only have $h(x) \equiv 0$. As has been pointed out, if the limits $a, b$ are fixed then you can't say anything about $h(x)$ over the range $a \le x \le b$, all you know is that it the integral vanishes and that $h(x)$ has some positive bits that cancel out with some negative bits.
A: If the integral of two functions are equal to each other, informally, you can use this to say that they have the same "shape." As in, if you subtracted one or the other by a certain constant, they would overlap completely. This is seen when integrating 1/sqrt(1-x^2) and -1/sqrt(1-x^2), both times with a trig substitution of sin(theta) = x. From a table, the second integral should yield arccos(x), but you can make it yield -arcsin(x) instead, which shows that -arcsin(x) and arccos(x), when taking a difference of two y values on their range, have the same shape. The distance between two points is the same, even if their values are different. Sorry for notation, haven't learned MathJax yet.
