If you integrate (I'd rather say, antidifferentiate) $f(x)$, you don't get $g(x)$, you get $g(x)+C$ for some arbitrary constant $C$. So it's wrong to think of $g(b)$ as "the area up to $x=b$."
Let's look at an example. When you antidifferentiate $\sin x$, you could get $-\cos x$, or you could get $-\cos x+42$ (or any of a number of other correct answers, but these two will do for now). So, the area up to $\pi/2$: is it $-\cos(\pi/2)$, which is $0$? or is it $42$?
Well, neither of these is a particularly good answer, which illustrates the difficulties you get into if you think of $g(b)$ as the area up to $x=b$.
Rather, you should think of $\int_a^xf(t)\,dt=g(x)-g(a)$ as the area from $a$ to $x$; it's only definite integrals that have an unambiguous interpretation as areas.
Notice that if we use a new symbol, say, $h(x)=\int_0^xf(t)\,dt$, for the area from zero to $x$ then if $f$ is always positive and $a$ is negative we'll have $h(a)$ negative, so in that sense area to the left of the $y$-axis is negative. But what this really shows is that you can't always interpret $\int_a^bf(x)\,dx$ as an area; you can always calculate the definite integral, but the number that comes out will not be what we generally accept as the area unless $a\le b$ and $f(x)\ge0$ for $a\le x\le b$.
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
While many mathematicians often like to forget about units, understanding units will be really helpful here.
Some transformations, such as $f(x) \mapsto f(x) + 3$, $f(x) \mapsto 4f(x)$, and $f(x) \mapsto f(x+3)$, are easy to understand visually looking at the graphs. Because the start function and the end function have the same units, it makes sense to plot them on top of one another on the same axes.
The transformation $f \mapsto f'$ is not like this. If $f$ has units of meters, and $x$ is seconds, then $f'$ has units of meters per second. This means that "comparing" them as you have done by overlaying them on top of each other is nonsense. For example, you can't glean anything from the fact that 5 meters per second is larger than 3 meters, or from the fact that 2 meters per second equals 2 meters.
The relationship between $f$ and $f'$ is not understood by comparing one directly to the other, but rather by understanding what features of $f$ correspond to features of $f'$. Draw $f$ and $f'$ on separate pairs of axes, and there are many features of import, many of which you probably know: for instance, zeros of $f'$ are local min- and maxima of $f$.
In short, the fundamental relationship between the two functions is not of a nature that you can just compare one to the other by overlaying their graphs. Your problem arises because you expect the graphs of $f$ and $f'$ to be visually comparable in an obvious way, which they are not. Instead, you must be willing to accept a different way of understanding the relationship than the one you are looking for. (See the answer here which has already been linked.) It would be like trying to understand how to multiply two numbers $x$ and $y$ by only looking at the values of $2^x$ and $2^y$.