TITLE: Confusion in fundamental theorem of calculus
QUESTION [10 upvotes]: In the wikipedia article shown below, it says $A(x)$ represents the area beneath the curve $y=f(x)$ between $0$ and $x$.

Then it says $f(x)=A'(x)$, i.e. $A(x)=\int f(x)dx$
which means $\int f(x)dx$ represents the area beneath the curve $y=f(x)$ between $0$ and $x$.
I think there is nothing wrong up to here.

Now if we put $f(x)=e^{x}$ then $A(x)=e^{x}$ and at $x=0$, $A(x)=e^{0}=1$
That is, the area $A(x)$ beneath the curve $y=f(x)$ between ($0$ and $x=0$) is $1$.
Now how can we get non-zero area when we find the area between same points (i.e. $0$ and $0$)

REPLY [5 votes]: Note that $f(x)=A'(x)$ does not follow $A(x)= \int f(x) d x$ since in here $A(x)$ is a specific function but $\int f(x) dx$ gives all the possible antiderivatives of $f$. For example, if $f(x)=2x$ then $\int f(x) dx=x^2+c$ for any $c$, where as $A(x)$ could be $x^2+1$ but cannot be $x^2+c$ for any $c \ne 1$.
Hence, from $f(x)=A'(x)$, you can only follow that if $F(x)$ is one of the antiderivative of $f$ then $$A(x)=F(x)+c=e^x+c. \tag{1}$$ 
Thus, you can't imply to $A(x)=e^x$ from $f(x)=A'(x)$. Instead, it is automatically understood that area between $0$ and $0$ is $0$, so that means $A(0)=0$. Hence, from (1) we follow $c=1$. Thus, $A(x)=e^x-1$ is the area between $0$ and $x$.