guys I figured out the stupidest fucking way to compute the area of a right triangle
ₐ₋₁
∫ f(x)dx
⁰
f(x) = ax
where a = the slope of the hypotenuse
comments
that doesn’t seem like that should be possible if you’re only using the slope, how does it take into account the scale of the triangle?
tbh I do not know how integrations work
the integral of a function is the area under the function’s curve between two x-axis bounds. A linear function is always a straight line. Limiting a linear function between 0 and a number n cuts off the portion of the line extending behind x = 0 and also cuts off the portion of the line extending beyond x = n. This creates a line segment with a slope of n - 1. Since the integral takes the area between the function and the x-axis between two bounds, it essentially draws two more lines: one from (0, 0) to (n, 0) and another from (n, (n - 1) n) to (n, 0). This creates a right triangle with a height of (n - 1) * n and a base of n. Since the integral takes the area of a function between two bounds, it takes the area of this invisible triangle by taking the area of the graph portion under the hypotenuse of the triangle. This is equal to the area of the triangle.