Proof of y'=f'(x): Showing g(x)=f(x)+constant

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The discussion centers on proving that if y' = f'(x), then any integral g(x) of f'(x) can differ from f(x) by at most a constant. The key steps involve defining w as the difference between f(x) and g(x), leading to w' = f'(x) - g'(x) = 0, which implies w is a constant function. This is supported by the Mean Value Theorem, confirming that if two functions have the same derivative, they differ by a constant, thus establishing that g(x) = f(x) + constant.

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Homework Statement


If y' = f'(x)
Then y = f(x) in one integral.
Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). Show that g(x) can differ from f(x) by at most a constant
Then Let : w' = f'(x) - g'(x) = 0
Then the equation of w as a function of x must be w = constant
hence we see that w = f(x) - g(x) equals g(x) = f(x) + constant.
Because g(x) is any integral other than f(x) all integrals are given by y = f(x) + c



How did they get from w = constant to g(x) = f(x) + constant algebraically?
second... they said Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). But if they are two dif integrals why did they set g'(x) = f'(x)? Doesn't setting something equal to something mean they are equal? Is there something that I am misunderstanding?
 
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Try looking at the mean value theorem.
 


Miike012 said:

Homework Statement


If y' = f'(x)
Then y = f(x) in one integral.
Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). Show that g(x) can differ from f(x) by at most a constant
Then Let : w' = f'(x) - g'(x) = 0
Then the equation of w as a function of x must be w = constant
hence we see that w = f(x) - g(x) equals g(x) = f(x) + constant.
Because g(x) is any integral other than f(x) all integrals are given by y = f(x) + c



How did they get from w = constant to g(x) = f(x) + constant algebraically?
There is a line missing- did you copy word for word? Rather than saying "let w'= f'- g'" you (they?) should say "let w= f- g". Then it follows that w'= f'- g'= 0. And from that "w= constant". Now we have w= f- g= constant so, adding g to both sides, f= g+ constant.

The fact that "w= constant" follows, as hunt_mat said, from the mean value theorem:
If w is continuous and differentiable, then (w(a)- w(b))/(a- b)= w'(c) where c is some value of x between a and b. In particular, if w' is identically 0, we have w(a)- w(b)= 0(a- b)= 0 so that w(a)= w(b). Since a and b can be any values of x, it follows that w is a constant function.

second... they said Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). But if they are two dif integrals why did they set g'(x) = f'(x)? Doesn't setting something equal to something mean they are equal? Is there something that I am misunderstanding?
Yes, setting them equal to something means they are equal to each other- g'(x)= f'(x). Which follows from the fact that both are anti-derivatives of f'(x).

But it does NOT follow that if f'(x)= g'(x), then f(x)= g(x). That is the whole point of this proof.
 

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