How is the slope of one function related to area under the graph of another function? So far, I believe I have a clear understanding of integrals and derivatives. If I am not wrong, when we integrate a function(Let's say $f(x)$), we find its anti-derivative(say $F(x)$). But what I fail to understand is the logical/mathematical relation between and integral and a derivative.
In other words, How is the derivative of one function giving us the integral of another? What exactly is the relationship?
Let us take a simple example of $f(x) = 3x^2$,whose indefinite integral is $x^3$. So why is $x^3$'s derivative giving us $3x^2$'s integral?
 A: Well, if we assume $f$ to be continuous, the relationship between $f$ and its integral, $F$, is given by the Fundamental Theorem of Calculus, which asserts that, if $f:[a,b]\to\mathbb R$ is continuous, then $F:[a,b]:\to\mathbb R$ given by
$$F(x) = \int_a^x f(t) dt
$$
is differentiable and $F'(c) = f(c)$ for all $c\in[a,b]$. We can easily read this result as on an interval where $f$ is continuous, the area between the curve of $f$ ($F$) and the horizontal axis grows/decreases at a rate described by $f$.
This is the reason why we say that $F$ is the primitive of $f$.
A: The derivative doesn't give you the integral of an other function.  The derivative gives you a function that has the first function as an integral.
The derivative of $x^3$ is not $3x^2$'s integral.  It is $3x^2$.  It's that$3x^2$'s integral is the original $x^3$.
The relationship is if:
$F'(x) = f(x)$ then  $\int f(x) dx = F(x)$.
basically (and theres a lot of technicalities I'm ignoring about whether derivatives/integrals exist and the whole "plus a constance" equivalence between classes of integers) derivatives and integrals are inverses.
$(\int f(x) dx)' = f(x)$ and $\int f'(x) dx = f(x) + C$.
If $F(x) = x^3$ and $f(x) =3x^2$ then $F'(x) = f(x)$  (not $\int f(x) dx$ as you said several times)  and $\int f(x) dx = F(x) + C$.
Why?  Well that's another question.
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We'll the derivative of $x^3$ is the integral of another function.  $(x^3)'  = \int 6x dx$.
That is, the derivative is the integral of the double derivative.
