Deriving Big Five Equations 3 and 4 from a Graph

The equations number 3 and 4 of the big five are
d = v1t + 1/2at^2
d = v2t - 1/2at^2
But there question is that where did these equations come from. Certainly we can't just accept everything as it is. We known already that to find the displacement of a section of a d-t graph we simply find the area between the graph and the x-axis.

The the graph above the area below the first vertical line can be seen as a trapezoid. While we can find its area using a simple equation, it would be easier to break it down into a triangle and a square.
One way of finding out the total area is to find the area of the triangle and the square individually and add them together.
The area of the square is v1t since its initial velocity can be seen as the length and the time as the width. The area of the triangle is 1/2(v2-v1)t as the height is v2-v1 and t once again acts as the base. Adding then together, we get d=v1t + 1/2(v2-v1)t. If we subsitute v2-v1=at then we get d=v1t + 1/2at^2 which is the 3rd equation.
The other way to find the area is to see it as one large rectangle and then subtract the triangle from it. The area of the large rectangle is v2t and we can reuse the area of the triangle from before. After subtracting one from the other we get d=v2t - 1/2at^2 which is the fourth equation.