Can't understand Second Fundamental Theorem of Calculus sorry if this has been asked before, but I can't seem to find my question in particular.
Anyway, in the Second FTM it says $$F(x)=\int_a^xf(x)dx$$
If I understand correctly is just the area under the curve. no problem there.
Then it says that$$\int_a^bf(x)dx=F(b)-F(a)$$
if I'm thinking this correctly, it would make sense because in  $F(a)=\int_a^af(x)dx$
 the area is $0$ so I'm just left with $F(b)$ which is the integral from a to b, this is where I think I have to be wrong, because in every example I see, they take the value of $a$ and plug it in $F(x)$.
For example in $$\int_2^5x^2dx$$ with $F(x)=x^3/3$
 they take $F(5)$ and $F(2)$, in particular $F(2)=2^3/3$, but shouldn't $F(2)$ always be $0$ because it's basically just $\int_2^2x^2dx$?
p.s Sorry in advance for any mistake I made in formating or anything else.
 A: 
Here are some aspects which might be helpful
  
  
*
  
*Be careful about not mixing up in the first part of FTC the variable of integration with the variable $x$. 
  \begin{align*}
F(x)=\int_a^xf(t)\,dt\tag{1}
\end{align*}
  The variable $t$ in (1) is a bound variable similar as the index $k$ in a sum $\sum_{k=0}^nf(k)$, while $x$ is a free variable.
  
*The function $F$ is not the area under the curve. The area is a number while $F$ is not a number but a function in $x$.
But, when we evaluate the function $F$ at a specific value $x=b$ we obtain the number
  \begin{align*}
F(b)=\int_{a}^bf(t)\,dt+F(a)
\end{align*}
  resp.
  \begin{align*}
\int_{a}^bf(t)\,dt=F(b)-F(a)\tag{2}
\end{align*}
  and $F(b)-F(a)$ in (2) can be interpreted as (signed) area under the curve in $[a,b]$.

A: This kind of confusion is very prevalent and the primary reason behind the confusion is the wrong definition of definite integral as area under a curve.
It is important to first understand that the definition of symbol $\int_{a}^{b}f(x)\,dx$ has nothing do with area as such. The definition has to be based on numbers $a, b$ and the function $f$ defined on interval $[a, b]$. One such definition was provided by Bernhard Riemann and it assumes $f$ to be bounded on $[a, b]$. I will leave the definition of Riemann integral to the standard textbooks of analysis and focus next on the Fundamental Theorems of Calculus.
First Fundamental Theorem of Calculus: If $f$ is bounded on $[a, b]$ and the Riemann integral $\int_{a}^{b}f(x)\,dx$ exists then the function $F$ defined on $[a, b]$ by $$F(x) = \int_{a}^{x}f(t)\,dt$$ is continuous on $[a, b]$ and $$F'(c) = f(c)$$ for any point $c \in [a, b]$ where $f$ is continuous.
Second Fundamental Theorem of Calculus: If $F$ is differentiable on $[a, b]$ and the derivative $F' = f$ (say) is Riemann integrable on $[a, b]$ then $$F(b) - F(a) = \int_{a}^{b}F'(x)\,dx = \int_{a}^{b}f(x)\,dx$$
Note that the first FTC describes a function $F$ based on a given function $f$ which has some nice properties ($F$ is continuous on $[a, b]$ and differentiable at those points where $f$ is continuous). But this $F$ is not an anti-derivative of $f$ on $[a, b] $ because we are not guaranteed that $F'(x) = f(x)$ for all $x \in [a, b]$. We have $F'(x) = f(x)$ only at those points $x$ at which $f$ is continuous. At points where $f$ is discontinuous the function $F$ may or may not be differentiable.
The second FTC deals with anti-derivatives. It says that if $F$ is anti-derivative of $f$ on $[a, b] $ i.e. $F'(x) = f(x)$ for all $x \in [a, b]$ and further that if $f$ is Riemann integrable on $[a, b]$ then the definite integral $\int_{a}^{b}f(x)\,dx$ can be simply evaluated in terms of the difference between values of anti-derivative (i.e. as $F(b) - F(a)$).
The function $F$ used in first FTC has a different role than the function $F$ of the second FTC and it is useless to think of them as same. Things change drastically when the function $f$ (which is to be integrated) is guaranteed to be continuous on $[a, b]$. When $f$ is continuous then both the FTC get merged into one theorem which we can simply call FTC for continuous functions:
Fundamental Theorem of Calculus for Continuous Functions: If $f$ is continuous on $[a, b]$ then $\int_{a}^{b}f(x)\,dx$ exists and the function $F$ given by $$F(x) = \int_{a}^{x}f(t)\,dt$$ is an anti-derivative of $f$ on $[a, b] $. Moreover if $G$ is any anti-derivative of $f$ on $[a, b]$ then $$\int_{a}^{b}f(x)\,dx = G(b) - G(a)$$
And now you see the connection between the word "anti-derivative" and the integral $$\int_{a}^{x}f(t)\,dt = F(x)$$ The function $F$ is an anti-derivative of $f$ and not necessarily the anti-derivative of $f$. You also remember the fact that a function does not have a unique anti-derivative and two anti-derivatives of the same function differ by a constant and this is the reason for using the constant of integration while calculating indefinite integrals. This is also the reason that I have used a different letter $G$ for a generic anti-derivative in the theorem mentioned above and the letter $F$ denotes a very specific anti-derivative represented in the form of a definite integral.
Thus when you wish to calculate $\int_{2}^{5}x^{2}\,dx$ by the use of anti-derivative $x^{3}/3$ you are just choosing one of the infinitely many anti-derivatives available and it will work as expected. You may wish to choose the specific anti-derivative $\int_{2}^{x}f(t)\,dt = (x^{3}/3) - 8/3$ and this will also work fine.
You can also think in this manner. The anti-derivative $F(x) = \int_{a}^{x}f(t)\,dt$ is a very specific one and has the property $F(a) = 0$. Other anti-derivatives of $f$ will not have this property of vanishing at $a$. And the fundamental theorem says that the choice of anti-derivatives does not matter in evaluating the definite integral.
A: There seems to have been a mix-up in what the letter $F$ refers to in each case.
What the following equation means
$$F(x) = \int_a^x f(x)\, dx$$
is that $F$ is a particular choice of antiderivative of $f$; that is, $F$ is a function such that $F'=f$, and moreover $F(a)=0$ for this choice of antiderivative.
However, the following equation
$$\int_a^b f(x)\, dx = F(b)-F(a)$$
is true for any choice of antiderivative $F$; that is, any function $F$ for which $F'=f$ will suffice. Antiderivatives are only defined up to a constant of integration, but the constant of integration cancels out since you're subtracting one from another.
In the first of these equations $F$ is a specific choice of antiderivative; whereas in the second equation, the constant of integration can be anything.
To resolve your issue in the case where $a=2$, $b=5$ and $f(x)=x^2$, note that
$$\int_2^x f(x)\, dx = \frac{x^3}{3} - \frac{8}{3}$$
so in this case, choosing $F(x) = \frac{x^3}{3} - \frac{8}{3}$, rather than simply $\frac{x^3}{3}$, gives you what you want.
