For constant velocity, one could directly prove area of a rectangle whose length is time traveled and height the velocity, is actually the displacement.
I would like to extend the same area logic for constant acceleration. However, to prove that area of triangle (as its a linear path due to increasing steady velocity), is the displacement, I need to prove, that summation of smaller rectangles contained in them add up to area of triangle (each rectangle is a displacement in a time period, so adding them all give total displacement).
But I could not prove these rectangles' area add up to the triangle as I could not connect it to summation and then integral formulas. Can anyone please help to prove the area of triangle via summation and then integration so I can connect all the dots?
You can write it as a summation, splitting it into equal amounts of rectangles:
$$\sum_{i=1}^n A_i$$
Where $A_i$ denotes the area of the $i$th rectangle. And the next step is to try to create an infinite amount of rectangles to essentially give us the area under the graph:
$$\lim_{n\to\infty} \sum_{i=0}^n A_i$$
And to find $A_i$ or the area of each rectangle, we multiply the width and the height of each rectangle. The height would be the height of the graph at each $i$th subinterval (thus $X_{ti}$ in your illustration), and the width of each would need to be $t_n-t_0$ divided by $n$ subintervals. This leaves us with:
$$\lim_{n\to\infty} \sum_{i=0}^n X_{ti}\left({t_n-t_0 \over n}\right)$$
You may notice this is in the form of an infinite Riemann sum, which can be rewritten as an integral:
$$\int_{t_0}^{t_n} f(x)\,\mathrm dx$$
Where $f(x)$ denotes the velocity graph. And if you want to go further and derive the formula for a triangle $\frac 12 bh$ via this integral, you can do like so. Assume that you have a linear velocity graph, with slope $\lambda$. That means:
$$f(x)=\lambda x \quad \lambda = {f(t_n)-f(t_0)\over t_n - t_0}$$ $$\begin{align}\int_{t_0}^{t_n} f(x) \,\mathrm dx &= \int_{t_0}^{t_n} \lambda x \,\mathrm dx \\ &= \frac 12 \lambda x^2 \Big|^{x=t_n}_{x=t_0} \\ &= \frac 12 \cdot {f(t_n)-f(t_0)\over t_n-t_0} \cdot \left({t_n}^2 - {t_0}^2\right) \\ \end{align}$$
Since in this case $t_0=f(t_0)=0$, we can simplify to this:
$$\begin{align} \int_{t_0}^{t_n} f(x) \,\mathrm dx &= \frac 12 \cdot {f(t_n)\over t_n} \cdot {t_n}^2 \\ &= \frac 12 \cdot f(t_n) \cdot {t_n} \end{align}$$
Which is $\frac 12 bh$.