we usually don't sum all the approximating rectangles, but perform that task indirectly by invoking the fundamental theorem of calculus.
we suppose that $F(t)=\int_a^t f(x)dx$ represents the area under the curve (or more correctly, the net area between the curve and the x-axis) of $f$ where $a\le x\le t$.
Now consider the derivative of A (the function whose values represent the slope of $A$) $F'(t)=\lim_{h\to 0} \frac{F(t+h)-F(t)}{h}$ $F'(t)=\lim_{h\to 0} \frac{\int_a^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$ $F'(t)=\lim_{h\to 0} \frac{\int_a^{t} f(t)dt+\int_t^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$ $F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$

Now think about what would happen to the shaded region to the right (between $t$ and $t+h$) were $h$ to become very small. The left and right sides of the region would become closer to each other in length and the region would resemble a rectangle with a width of $h$ and a height of $f(t)$. thus $F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$ $F'(t)=\lim_{h\to 0} \frac{hf(t)}{h}$ $F'(t)=f(t)$ Put another way, we say that the area function ($F$) is some anti-derivative of $f$.
For example, if we know that the antidervivative of $x^2$ is $x^3/3+C$ (we have an unknown constant because parallel curve have the same slope), we know that the area between where x ranges between $1$ and $2$ would be $2^3/3+C$, which is a pretty useless result. To fix the problem we note that area of a region is just the sum of the areas of its components.

In the second graphic, the total area under $f$ between $c$ and $b$ is just the sum of the blue and green areas. In the language of integrals, $\int_c^bf(x)dx = \int_c^af(x)dx + \int_a^bf(x)dx$ thus $\int_a^bf(x)dx = \int_c^bf(x)dx - \int_c^af(x)dx$ So to solve the area problem, we need to evaluate 2 anti-derivatives, not one. In our example $\int_1^2 x^2dx = (2^3/3+C)-(1^3/3+C)=(8-1)/3=7/3$ Notice how nicely the $C$s cancel out.
By the way, in single variable calculus, you can think of $dx$ as an infinitely small non-zero positive $\Delta x$. Note that $d$ and $\Delta$ are the Latin and Greek representations for the first sound in the word difference.