CALCULUS II True/False question NEED HELP!!

[NickMusic]

TRUE OR FALSE? If F(x)and G(x) are antiderivatives of f(x), then F(x)=G(x)+C.

I've been trying to do this, and I think it's true, but I'm not sure and I also have to explain why and I'm having a really hard time

Could someone help me please?!

 

[stapel]

TRUE OR FALSE? If F(x)and G(x) are antiderivatives of f(x), then F(x)=G(x)+C.

What is the definition (from your book or your class notes) of the antiderivative of a function? :wink:

 

[NickMusic]

What is the definition (from your book or your class notes) of the antiderivative of a function? :wink:

It says , A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I.

Informally, F(x) is the general antiderivative of f(x) if (dF/dx)= f(x)

But then I don't understand the G(x) ... like what does he have to do with this?

 

[JeffM]

It says , A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I.

Informally, F(x) is the general antiderivative of f(x) if (dF/dx)= f(x)

But then I don't understand the G(x) ... like what does he have to do with this?

\(\displaystyle U(x) = (x + 1)(x + 2) \implies U'(x) = (x + 1)(1) + (x + 2)(1) = x + 1 + x + 2 = 2x + 3.\) Correct?

\(\displaystyle V(x) = x^2 + 3x - 25 \implies V'(x) = 2x + 3.\) Right?

\(\displaystyle So\ U'(x) = V'(x) \implies U(x) = V(x) + C\ because\ (x + 1)(x + 2) = x^2 + 3x + 2 = (x^2 + 3x - 25) + 27.\)

Antiderivatives are unique EXCEPT for a constant term.

 

[stapel]

It says , A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I.

But then I don't understand the G(x) ... like what does he have to do with this?

G(x) is just the name they've given of another function which is an antiderivative of f. It's just the function notation you've been working with since back in algebra.

Since F' = f and G' = f, then F' = G', right? So F' - G' =... What?

What is the antiderivative of that value? And what is the antiderivative of F' - G'?

Solving for "F=", what do you get? :wink:

 

[HallsofIvy]

To prove that, you really need the "mean value theorem": if f is continous on [a, b] and differentiable on (a, b) then there exist c between a and b such that \(\displaystyle \frac{f(b)- f(a)}{b- a}= f'(x)\).

If F'(x)=f(x) and and G'(x)= f(x), let H(x)= F(x)- G(x). Then H'(x)= F'(x)- G'(x)= f(x)- f(x)= 0 for all x. Therefore, by the mean value theorem, for any a and b, \(\displaystyle \frac{H(b)- H(a)}{b- a}= 0\) so that \(\displaystyle H(b)- H(a)= 0\) and so H(b)= H(a). That is, any function, whose derivative is 0 for all x, is a constant. H(x)= F(x)- G(x)= C so F(x)= G(x)+ C.