Suppose someone came and told you:
"Hey, I've integrated a function $f(t)$ over the interval $[a,x]$ and I have got the following
$\int_a^x f(t) dt =F(x)$
I know an explicit formula for $F(x)$ [enter your choice of $F$ here], but sadly I have forgotten what $f$ was. What can I do?"
Then you can say "Use the Fundamental Theorem of Calculus. That will tell you that
$f(x)=F'(x)$
and you're done!"
But the guy might be curious and ask, "Hey, but how does that work?!"
For the explanation, check this
The importance of the FTC is that it connects the notion of integral and derivative, which wasn't something obvious back then. FTC is saying that the process of integration and the process of integration are inverses of each other. In fact, if we define two sets
$A=\text{the set of all continuous functions on }[a,b] \;\,,f:[a,b]\to \Bbb R$
and $C\subset A$
$C=\text{the set of all continuous functions, with continuous derivative on }[a,b] \text{ and } f(a)=0$
Then the functions $I$ and $D$ defined as
$I:A\to C$ such that
$I(f)(x)=\int_a^xf(t)dt\,\; ;a\leq x \leq b$
and
$D : C \to A$ such that
$D(f)(x)=f'(x)$
are in fact inverse mappings.
Note that
$(I \circ D(f))(x)=\int_a^xf'(t)dt=f(x)-f(a)=f(x)$
since $f(a)=0$, and
$(D \circ I(f))(x)=\frac{d}{dx} \left(\int_a^xf(t)dt\right)=f(x)$
You can see the "proof" that they are indeed inverse mappings relies wholely on the FTC! Note that $D$ and $I$ take a function $f$ as an input and output a function $I(f)$ or $D(f)$. The fact we write $(x)$ next to them is to remind they are still functions of $x$.
ADD: For the practical uses, FTC is the most efficient tool to evaluate definite integrals. When definite integrals are defined, we can use lower and upper sums, Riemann sums and such to find some of them, such as
$\eqalign{ & \int\limits_a^b {{x^p}dx} = \frac{{{b^{p + 1}} - {a^{p + 1}}}}{{p + 1}} \cr & \int\limits_a^b {{e^x}dx} = {e^b} - {e^a} \cr} $
but other integrals become more and more laborious. The FTC allows us to evaluate integrals such as
$\int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 + {x^2}} }}} $
Surely, there are far more interesting examples, but I can't come up with one now.