1 Aug
2023
1 Aug
'23
12:53 p.m.
i think i did integral wrong. that it's not 3x^3 but rather 1/3 x^3 . seems quite likely. let's try to find via another route. ummmmmmmmmmmmm thinking of the 1/2 in the triangle area, since we are trusting elsewhere. is this related to 1/2 x^2 somewhere maybe? or 1/2 x? the area of a triangle is integral of y(x) = slope * x where slope = h/w and integral is 0...w so integral(x * h / w, dx, 0...w) = 1/2 h * w uhh indefinite integral, assuming exponent goes to denominator 1/2 x^2 * h / w | x = w 1/2 w^2 * h / w = 1/2 w * h so integration puts the _new_ exponent in the _denominator_. check with calculating triangle area.