on some definite integral problems that Sf(x)dx i use [f^2(x)]/[2*f'(x)] and it gives me the correct answer but it gives me bad vibes idk why. i know its wrong because the derivative of that function isnt f(x) but it always works out. are there any nerds who can clarify why this works? (f(x) is polynomial)
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f:R->R, g(x)=2x+2
-1/int/+1[f(g(x))]dx = 18
2/int/4[g(f(x))]dx = 18
0/int/2[f(x)]dx=?
I tried to see what you did and I got this suspicion that you are doing 1st degree polynomials.
It's just a funny coincidence. When you plug a standard 1st degree polynomial in the formula you provided you get ax^2/2 + bx + b^2/2a. This is very close to what you get by standard integration of a 1st degree polynomial (ax^2 + bx + constant).
Because you are doing definite integrals, that b^2/2a term disappears and you remain with just the ax^2/2 + bx. This is just a coincidence. I don't think it works for higher degree polynomials. (I'm too lazy to check lol)
TLDR: coincidence
im pretty sure youre right. its just weird that this has worked for all the problems so far. crazy luck
You're right I think. The derivative of this thing is basically f(x) + f''(x) * (...). So if the second derivative is zero this is actually exact. Tbh I'd say it's a pretty okay approximation if the first derivative is large too since the second term is divided by f'(x)^2.
You're using [f^2(x)]/[2f'(x)] as a u-substitution, and it only works out properly because f(x) is a linear polynomial. The moment f(x) is a a higher degree polynomial (like if f(x) = x^2) it doesn't work because the substitution is essentially a reverse engineering of how it works for linear polynomials. f(x) is only the derivative of [f^2(x)]/[2f'(x)] if f'(x) is constant.
What #7 said is right. It only works with linear, and essentially is just a big coincidence. Sorry if this sounds like a word jumble i suck at teaching Lmao
seems like i was late, but you already got your explanation (#5 is right, its just that the constant term gets elimininated in definite integration)
i am interested about how did you stumble upon this substitution? By chance or experimenting with different options and this just worked
essentially i just thought what f(x) is derived from. just like how x is x^2/2 i did f^2/2 but when you derive f^2 there is also an f'x so i put that in the denominator too. it IS wrong but its an easy way to solve first degree polynomials integral i guess. a timesaver
tldr d/dx(f^2(x))= 2f(x)*f'(x) => intf(x)dx = f^2(x)/2f'(x)
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