by Jhdoe
Last Updated February 11, 2019 10:20 AM

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 ?

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 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: $\frac{x^3}{3}+\frac{x^2}{2}+C$!

$$ \int g(x)\,dx=\int[x(x+1)]\,dx=\int (x^2+x)\,dx=\int f(x)\,dx=\frac{x^3}{3}+\frac{x^2}{2}+C. $$

P.S. Don't confuse two different concepts: the definite integral and the indefinite integral. They're different! There is something that unites them, however, and that something is called the fundamental theorem of calculus.

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

$$F(x_0)+C_F=G(x_0)+C_G.$$

Now it holds for all $x$.

In other words, subtracting both,

$$F(x)-F(x_0)=G(x)-G(x_0)$$

or

$$\int_{x_0}^x f(t)\,dt=\int_{x_0}^x g(t)\,dt.$$

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