Important: the derivative has a geometrical interpretation as you state, but that doesn't mean it is the definition of the derivative.
Try and see it this way: when we integrate (in Riemann's way), we are making a "continuous sum" of a continuous function $f(x)$ with respect to the infinitesimal quantity $dx$ over an interval $(a,x)$ - we are taking the "product" of $f$ and $dx$ over the infinitely many real values in $(a,x)$ and "summing" them up, getting a new function $F(x)$.
That is, suppose we define
$$F(x) = \int\limits_a^x f(t) dt$$
Then we can put this geometrically in the following diagram:
We're interested in finding $F'(x)$, so we construct our difference:
$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{F\left( {x + \Delta x} \right) - F\left( x \right)}}{{\Delta x}}$$
We can write that expression as
$$\frac{1}{{\Delta x}}\left( {\int\limits_a^{x + \Delta x} {f\left( t \right)dt} - \int\limits_a^x {f\left( t \right)dt} } \right)$$
But using some theorems from definite integration we have
$$\frac{1}{{\Delta x}}\left( {\int\limits_x^{x + \Delta x} {f\left( t \right)dt} + \int\limits_a^x {f\left( t \right)dt} - \int\limits_a^x {f\left( t \right)dt} } \right) = \frac{1}{{\Delta x}}\int\limits_x^{x + \Delta x} {f\left( t \right)dt} $$
We also now that if $f(x)$ is continuous (which we assumed is), we have
$$\frac{1}{{\Delta x}}\int\limits_x^{x + \Delta x} {f\left( t \right)dt} = \frac{1}{{\Delta x}}f\left( \xi \right)\Delta x = f\left( \xi \right)$$
for some $\xi$ in $[x,x+\Delta x]$
Now taking the limit produces
$$F'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{F\left( {x + \Delta x} \right) - F\left( x \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} f\left( \xi \right)$$
But we have that $\xi \in \left[ {x,x + \Delta x} \right]$, which means that
$$\mathop {\lim }\limits_{\Delta x \to 0} f\left( \xi \right) = f\left( x \right)$$
What have we done? We have retrieved the original function $f$ which we summed along with $dx$ over $(a,x)$ to obtain $F(x)$. So what does this tells us? That differentiation in the "operational" sense, reverts the process of integration, just like multiplication "reverts" the process of division.
I'm not a tacher or tutor or anything of the sort, so maybe you can get better answers from such people, but I hope you understand what I intended to explain.
Another imporant consequence of this is the following:
We have proven that if we have a function $f(x)$ and defined $F(x)$ as
$$F\left( x \right) = \int\limits_a^x {f\left( t \right)dt} $$
then $F'(x) = f(x)$, that is, $F$ is a primitive of $f$. But we know that two primitives of $f$ will only differ by a constant term, so, let $G$ be another primitive. We have that
$$\int\limits_a^x {f\left( t \right)dt} - G\left( x \right) = C$$
But then putting $x=a$ gives
$$ - G\left( a \right) = C$$
We get a new result. If $G$ is a primitive of $f$, then the following holds:
$$\int\limits_a^x {f\left( t \right)dt} = G\left( x \right) - G\left( a \right)$$
Well...it's not obvious, as you point out, why "finding a slope" and "finding area under a curve" are opposites. That non-obviousness is why it's called a "theorem" (indeed, more formally stated, it's called "the fundamental theorem of calculus").
But maybe I can help out with the intuition a little. Instead of thinking of $f'(b)$ as the slope of the graph of $f$ at the point $(b, f(b))$, think of it this way: if we moved a little to the right of $b$, say, to $b + h$, where $h$ is a small number, what would we expect $f(b+h)$ to be? Well, if you draw a picture, you'll see that we expect it to be
$$
f(b+h) \approx f(b) + h * slope
$$
and since the slope is $f'(b)$, this is
$$
f(b + h) \approx f(b) + h f'(b).
$$
That's true for any "nice" (smooth, etc.) function, for small values of $h$. Let me define such a function, the "accumulated area" function:
$$
F(x) = \int_0^x f(t) ~dt.
$$
That's the area under the graph of $f$ between $0$ and $x$. In particular, we have
$$
F(b) = \int_0^b f(t) ~dt
$$
is the area under the graph of $f$ from $0$ to $b$. What's the value of $F(b + h)$ for a small value of $h$?
We have two possible answers: the first is from the definition: it's just
$$
F(b+h) = \int_0^{b+h} f(t) ~dt = \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt.
$$
Hold that thought.
The second answer comes from that general stuff I wrote above, namely:
$$
F(b+h) \approx F(b) + h F'(b).
$$
Setting these two equal (since they both represent the value of $F(b+h)$, more or less), we get
$$
F(b) + h F'(b) \approx \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt
$$
Since the $F(b)$ on the left hand side is the same as the first integral on the right hand side, this simplifies to
$$
h F'(b) \approx \int_b^{b+h} f(t) ~dt
$$
and dividing through by $h$, becomes
$$
F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt
$$
Now when $h$ is very small, the integrand, $f(t)$ is roughly constant ... and its value is approximately $f(b)$, So the integral on the right is roughly the integral of the constant $f(b)$ over an interval of width $h$; that value is just $h f(b)$. So we get
$$
F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt \approx
\frac{1}{h} [h f(b) ] = f(b).
$$
So the derivative of the "accumulated area" function $F$ is the original function $f$.
As for derivative and integral being "opposites", you might want to look at
$$
f(x) = x^2 + 1.
$$
Try taking its derivative, to get a new function $g$, and then write down the accumulated area function
$$
G(x) = \int_0^x g(t) dt.
$$
Evaluate the integral, and see whether it equals $f$. (Hint: it doesn't!). So "differentiate then integrate" doesn't necessarily bring you back to the original function. On the other hand, for nice enough functions (e.g., continuous), "integrate then differentiate" does bring you back to the original.
Best Answer
If you want to look at derivatrive as a map $\mathcal D:\mathcal F\to \mathcal F$ where $\mathcal F$ is the set of all (differentiable) functions, then $\mathcal D$ is neither one-one, nor onto. That's why, it's not okay to talk about inverse of $\mathcal D$ in the same sense as you would talk about inverse of $F$ where $F:\mathcal F\to \mathcal F$ and $F(f)=f+5$.
Since the derivative maps a set of functions (namely those which are separated only by a constant) to one, we need to look at the inverse also as a set of functions. I hope you know that if $f:D\to R$ is not an onto function, then we can (and do) occasionally define $f^{-1}(y)=\{x:f(x)=y\}$. You can think of a similar case here.