After reading some of the comments and from outside help I've been made aware that my notation may have been the problem. So I'm rewriting this with a new set of questions with better notation and hopefully better understanding.
Let me start from the very beginning – Limits: $\lim_{x\to a} x$ refers to making x very small and almost $a$ but $x\ne a$
The next thing is derivation. Derivative of $f(x)$ if the limit exists refers to the new function $f'(x)$ whose each input $x$ maps to $f$ and output is the slope at that point in the $f$. In other words, the ratio of infinitely small change in $y$ when infinitely small change happens to input $x$ for every point. Mathematically,
$$
\lim_{\triangle x \to 0}\frac{f(x + \triangle x) – f(x)}{\triangle x}
$$
Indefinite integral in simpler words is just the opposite of derivative. If the derivative of $f(x)$ is $f'(x)$ then the integral of $f'(x)$ is $f(x) + C$ where $C$ can be any constant which is shown by pushing the curve up or down the slope doesn't change so all $C$ would make the derivative same as $f'(x)$
$$
\int f'(x) = f(x) + C \ \ \ \ \& \ \ \ \ \frac{\mathrm{d}f(x)}{\mathrm{d}x}
= f'(x)$$
Problem come in Definate Intregal. To my understanding Definite integral is integral between fixed range $[\alpha, \beta]$ given function is continuous. This solves our problem of $C$ constant as per how it is calculated. According to Theoram:
$$
\int_\alpha^\beta f'(x) = f(\beta) – f(\alpha)
$$
which makes sense as:
$$
\begin{align}
& \ \ \ \ \ (f(\beta) + C) – (f(\alpha) + C) \\
&= f(\beta) – f(\alpha) + (C-C) \\
&= f(\beta) – f(\alpha)
\end{align}
$$
Q1. But why is it $f(\beta) – f(\alpha)$ in the first place?
Now I would also want to talk about Area. To get the Area of the curve we divide it into little rectangles with the width of $\triangle x$ which tends to 0 bur $\triangle x \ne 0$
Mathematically area will be:
$$
A() = \lim_{\triangle x \to 0} \triangle x \sum_{i=\alpha}^\beta f(x_i + (i-1)\triangle x)
$$
Since from pic $\alpha = 0.5\ \ \beta = 2.5$ the above equation is:
$$
A() = \lim_{\triangle x \to 0} \triangle x \sum_{i=0.5}^{2.5} f(x_i + (i-1)\triangle x)
$$
This equation by definition of integration is:
$$
\int_{\alpha}^\beta f(x) \triangle x
$$
Now Fundamental Theorem of Calculus says
$$
\int_{\alpha}^\beta f(x) \triangle x = F(\beta) – F(\alpha),\ \ F'(x) = f(x)
$$
Q2 – 3:
Q2. What is the relation of Area to antiderivative function?
Q3. How one thus comes to the logic of Subtracting from a geometrical perspective, here above image.
Best Answer
Using the geometric interpretation, $\int_a^b f(z) dz$ is the total signed area between $f$ and the $x$-axis, where we show this intuition as the limit of Riemann sums.
Now, let's replace $b$ with a variable $x>a$ and ask what is the rate of change of the area under the curve at $x$:
$$\frac{d}{dx}\int_a^x f(z) dz = \lim_{h\to0^+} \frac{\int_a^{x+h} f(z) dz - \int_a^x f(z) dz}{h} = \lim_{h\to0^+} \frac{\int_x^{x+h} f(z) dz}{h}$$
$$=\lim_{h\to0^+} \frac1h\left[ \lim_{n\to \infty}\sum_{i=0}^n f\left(x+i\frac hn\right)\frac hn\right] = \lim_{h\to0^+}\left[ \lim_{n\to \infty}\frac1n\sum_{i=0}^n f\left(x+i\frac hn\right)\right]$$
Note that the inner sum is just the mean value of the terms in the sum and the inner limit defines the following Riemann integral:
$$\lim_{n\to \infty}\frac1n\sum_{i=0}^n f\left(x+i\frac hn\right) = \frac1h \int_x^{x+h}f(z)dz$$
Which is mean value of $f$ over $[x,x+h]$
In general if $f$ is integrable over $[a,b]$ then the function $F(x):= \int_a^x f(z)dz$ is uniformly continuous on $[a,b]$ and $F(a)=0$ so $\lim_{x \to a^+} F(x)=0$
Using this, we can switch the limits in the final sum:
$$\lim_{h\to0^+}\left[ \lim_{n\to \infty}\frac1n\sum_{i=0}^n f\left(x+i\frac hn\right)\right] = \lim_{n\to\infty}\left[ \lim_{h\to 0^+}\frac1n\sum_{i=0}^n f\left(x+i\frac hn\right)\right] $$ $$= \lim_{n\to\infty}\left[ \frac1n\sum_{i=0}^n f(x)\right]=\lim_{n\to\infty}f(x) = f(x)$$
So the rate of change in the area of the definite integral at $x$ is $f(x)$, as we expect.