How do integrals add up all the areas within their bounds? Before I proceed,
I am a high school student and have just started learning calculus.
OK
Now, I understand that an integral is basically the opposite of a derivative. I also understand that it finds the area under the graph of a function.
Based on what I understood, while integrating you take a very small change in the variable $x$ (or maybe $\mathrm{d}x$) such that the change in $f\left(x\right)$ is constant and therefore calculate the area of the rectangle and add the areas...
However, my question here is how does the integration tool include all the areas
bound under its limit, it makes less sense to me.
For example:
$$\int_2^4 x\,\mathrm{d}x$$
Solving the integral:
You get
$$\frac{x^2}{2}$$
with limits $4$ and $2$.
So, now you substitute the limits and subtract them .....
but this is what feels weird, it looks like I'm only subtracting the area at $4$ and $2$,
How does this tool work
(Any response would be helpful)
EDIT:
(i want to make my question a bit more clear)
[My question is basically, how does integration include all the areas between the interval, i have seen a couple of proofs on the fundamental theorem of calculus
however, that just seemed to explain why we subtract while integrating (to remove the constant term (i think).The problem i guess i am facing is that there is no hint of a summation of areas happening,and this is exactly what my question is -how do we know the areas are included??]
 A: First, you have to understand what a definite integral really is. First, you have to do is learn Riemann integral, you'll see $\int_{a}^{b} f(x)dx$ just denotes the area under the curve between the points a and b. As for your main question, why $\int_{a}^{b} f(x)dx=F(b)-F(a)$ where $F'(x)=f(x)$. You will need to learn Lagrange's theorem. I believe after that, you might be able to understand how the Fundamental Theorem of Calculus works
A: The way to understand this is to step back from thinking about integrals as areas.  Fundamentally, integrals are not areas, they are infinite sums of infinitely small values.  To understand how it works, let's look at the opposite direction - differentials.  Take the equation:
$$ y = x^2 $$
Now, let's differentiate it:
$$ dy = 2x\,dx $$
On the left, you have an infinitely small change in $y$.  On the right, you have infinitely small values, which are also equivalent to the same infinitesimal change in $y$.
Now, if you add up all of the infinitesimal changes in $y$ between two $x$ values, you will get the TOTAL change in $y$, correct?  Therefore, if you had the original function, you could just subtract the $y$ values at those points to get what the total change is.
A way of stating that mathematically is:
$$ \int_{x_0}^{x^1} dy = y(x_1) - y(x_0) $$
So that's what an integral is.  It is the sum of all of the infinitely small changes of an expression.  So how does it relate to areas?  Let's go back to your problem, where you want the area under the curve $y = x$.
With areas, an area of a rectangle is width times height.  So, each area will be $dx$ wide and $y$ high.  Therefore, each rectangle will be:
$$ y\,dx $$
Now, we can imagine a function $z$ such that its differential, $dz$, is the same value as this area.  So we set:
$$ dz = y\,dx $$
So, using the reasoning we used before, we know that the TOTAL of all of $dz$ is going to be equal to the difference between $z$ evaluated at $x_0$ and $x_1$.  In other words:
$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} $$
So, since $z(x_1) - z(x_0)$ is equal to the sum of the areas, so is $ \int y\,dx\, \biggr|_{x_0}^{x_1}$.
Now, we can't integrate $y\,dx$ directly, but, thankfully, we have a formula for $y$, which is $y = x$.  Therefore, this becomes:
$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} = \int x\,dx\, \biggr|_{x_0}^{x_1} = \frac{x^2}{2} \biggr|_{x_0}^{x_1} $$
So, what we did was:

*

*Recognize that the integral is an infinite sum of infinitely small pieces

*Recognize that the difference between two points on a function is equal to the infinite sum of the differentials of that function.

*Because of (2), integrating differentials is the same as performing an infinite sum of them.

*If we can write the formula for an infinitely small area as a differential, we can use (1)-(3) to generate a formula for a function such that the function's change is equal to the sum of the infinitely small values between those points.

