Integrating both sides of an equation Suppose i have
$g(x)=f(x)$
Now, i integrate both sides of the equation
$\int g(x)dx=\int f(x)dx$
I get:
$G(x)+C=F(x)+C$ , where G(x) and F(x) are anti-derivatives.
because G(x) is the area from 0 to x, as F(x), and because g(x)=f(x), then the 2 constants of integration, to me, must be the same. Is it correct ?
 A: Well, if you say that one function is equal to another function, it means that the result of integration, which is a family of functions that differ only by a constant, of one of the functions will be equal to the result of integration, which is again a family of functions that differ only by a constant, of the other one. That's obviously true because the functions are the same.
Let's say we have two functions $f(x)=x^2+x$ and $g(x)=x(x+1)$. I hope you can see that they are exactly the same function. So, we can write down the following: $f(x)=g(x)$. And we also know that $\int f(x)\,dx=\frac{x^3}{3}+\frac{x^2}{2}+C$. Then, what will $\int g(x)\,dx$ be equal to if we defined $g(x)$ to be equal to $f(x)$? Well, obviously it's going to be the same family of functions: $\frac{x^3}{3}+\frac{x^2}{2}+C$!
$$
\int g(x)\,dx=\int[x(x+1)]\,dx=\int (x^2+x)\,dx=\int f(x)\,dx=\frac{x^3}{3}+\frac{x^2}{2}+C.
$$
The notation $F(x)+C$ just means "a family of functions that differ only by a constant". It's really just a shorthand for this: $\{F(x)+C,C\in\mathbb{R}\}$.
P.S. Don't confuse two different concepts: the definite integral (antiderivative) and the indefinite integral (geometrically, the area under a curve). They're different! There is something that unites them, however, and that something is called the fundamental theorem of calculus.
A: Nearly, but as the antiderivatives $F$ and $G$ can themselves differ by a constant, the reasoning can fail.
What you can write is
$$F(x)+C_F=G(x)+C_G$$ and verify the condition at a single point, say
$$F(x_0)+C_F=G(x_0)+C_G,$$ which can be enforced with $$C_G=C_F+F(x_0)-G(x_0).$$
Now it holds for all $x$.
In other words, subtracting both, 
$$F(x)-F(x_0)=G(x)-G(x_0)$$
which is equivalent to 
$$\int_{x_0}^x f(t)\,dt=\int_{x_0}^x g(t)\,dt.$$
A: What you're computing is the indefinite integral and there's no reason for the constants of integration to be the same on both the sides. Let $\int f(x) \mathrm dx=F(x)+C_1$ and $\int g(x)\mathrm dx=G(x)+C_2$ In fact you may simply write the following:$$f(x)=g(x) \implies \int f(x)\mathrm dx=\int g(x) \mathrm dx\implies F(x)=G(x)+C \ \text{where} \ C=C_2-C_1$$Note that this is the case even during computation of definite integral, however the effect of the constants of integration gets nullified in the process:$$\int_{a}^{b}f(x)\mathrm dx=\bigl[F(x)+C_1\bigr]_{a}^{b}=F(b)-F(a)$$
A: Observe that if $f$ is differentiable on $\Bbb R$ then the derivative of both $f$ and $f+c$ are same which is $f'$ but $f \neq f+c$ unless $c = 0.$
