Prove that if $F(x)$ and $G(x)$ are antiderivatives of $f(x)$, then there exists a constant such that $F(x)-G(x)=C$.

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Lesson Parent: 
Antiderivatives
Problem: 

Prove that if $F(x)$ and $G(x)$ are antiderivatives of $f(x)$, then there exists a constant such that $F(x)-G(x)=C$.

Answer: 

It is true that if $F(x)$ and $G(x)$ are antiderivatives of $f(x)$, then there exists a constant such that $F(x)-G(x)=C$.

SAMPLE SOLUTION

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Solution: 

Since both <Sign in to see all the formulas> and <Sign in to see all the formulas> are equal to $f(x)$ we have

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Now since $-1$ is a constant and $G(x)$ is differentiable we know that

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Since, both $F(x)$ and $-G(x)$ are differentiable then we know that

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and since

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we know that

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Now let's call <Sign in to see all the formulas>. Because of our properties of function limits, we know that $H(x)$ is continuous on $[a,b]$. Similarly, from our differentiation formulas we know that <Sign in to see all the formulas> on $(a,b)$. From this problem we know that $H(x)$ is a constant. Let's call the constant $C$ so that <Sign in to see all the formulas> and therefore, <Sign in to see all the formulas>.

The proof is now complete.