Why does the area under a curve equal it's antiderivative's value at the upper limit? Based on what I've read online about calculus, it seems that the following assertion is true:
$$\int_{0}^{x}f' = f(x)$$
or i.e.
$$\int_{0}^{5}f' = f(5)$$
However, I do not really understand why.  I know that it is true only because it is said to be true, and because the values show it to be true.
Why is this true? I just can't really understand it properly in my head.  Why is the area under the graph of $f'$ equal to $f(x)$ when $x$ is equal to the upper limit of the definite integral?
 A: For a differentiable function $f$, you have the approximation 
$$
f(b)\sim f(a)+f'(a)\,(b-a).
$$
Then, by applying this to a partition $a=x_0 <\cdots < x_n=b$, we get
$$
f(b)-f(a)=\sum_{j=1}^nf(x_j)-f(x_{j-1})\sim\sum_{j=1}^nf'(x_{j-1})\,(x_j-x_{j-1}).
$$
The right-hand-side is a Riemann sum, so in the limit
$$
f(b)-f(a)=\int_a^b\, f'.
$$
A: Like David said, you need to subtract $f(0)$. Otherwise, you could reach a contradiction in the following way: let $f$ and $f'$ be as given in your question, and define $g(x) = f(x) + 1$. Then $g' = f'$, but if what you wrote is true,
$$
f(x) = \int_0^x f' = \int_0^x g' = g(x),
$$
a contradiction since we know that $g(x) = f(x) + 1$.
I think you might be getting confused here: the Fundamental Theorem of Calculus says that if you have a function $f$ that is continuous on $[a, b]$ and define $F$ by
$$
F(x) = \int_a^x f(t) dt,
$$
then $F'(x) = f(x)$ on $(a, b)$. However, there are other functions $G$ for which $G' = f$ on $(a, b)$ (namely, $F$ plus any nonzero constant) for which this equality does not hold; rather, the equality is, repeating again what David said,
$$
\int_a^x f(t) dt = G(x) - G(a).
$$
In fact, from how we defined $F$ above,
$$
F(a) = \int_a^a f(t)dt = 0,
$$
so that it is indeed the case that
$$
\int_a^x f(t)dt = F(x) = F(x) - 0 = F(x) - F(a).
$$
Anyway, if you're looking for some intuition about these results, let's return to the equality
$$
\int_a^x f'(t)dt = f(x) - f(a).
$$
Now, for each point $t$ in $(a, x)$, $f'(t)$ gives the signed rate of change of $f$ at $t$. So, intuitively, when we integrate the rate of change of $f$ ($f'$) over the interval $(a, x)$, we are in some way measuring the total change in $f$ over this interval; and the total change in $f$ over the interval $(a, x)$ is just $f(x) - f(a)$. To see this more clearly, if you're familiar with the definition of the average of a function $g$ over $(a, b)$ as
$$
\frac{1}{b-a} \int_a^b g(x)dx,
$$
we have that
$$
\int_a^x f'(t)dt = \frac{1}{x-a}\left[\int_a^x f'(t)dt\right] \cdot (x-a),
$$
which is to say, the integral of $f'$ over $(a, x)$ is just the average value of $f'$ on $(a, x)$ times the length of $(a, x)$; or, since $f'$ is just the rate of change of $f$, it is the average rate of change of $f$ on the interval $(a, x)$ times the length of the interval $(a, x)$, which gives us the total change of $f$ over the interval $(a, x)$: again, $f(x)-f(a)$.
Let me know if this doesn't answer your question or leaves you with more questions.
