Prove that if $F(x)$ is an antiderivative of $f(x)$ and $C$ is a constant, then $G(X)=F(x)+C$ is also an antiderivative of $f(x)$.

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Lesson Parent: 
Antiderivatives
Problem: 

Prove that if $F(x)$ is an antiderivative of $f(x)$ and $C$ is a constant, then $G(X)=F(x)+C$ is also an antiderivative of $f(x)$.

Answer: 

It is true that if $F(x)$ is an antiderivative of $f(x)$ and $C$ is a constant, then $G(X)=F(x)+C$ is also an antiderivative of $f(x)$.

SAMPLE SOLUTION

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Solution: 

Recall our definition of an antiderivative. It is a function that is continuous on $[a,b]$ and <Sign in to see all the formulas> for all $x$ in $(a,b)$. We must prove that $G(x)$ is continuous on $[a,b]$ and <Sign in to see all the formulas> for all $x$ in $(a,b)$.

First, let's prove that <Sign in to see all the formulas> for all $x$ in $(a,b)$. Recall that the derivative of a constant is zero and that the derivative of two differentiable functions is the sum of their derivatives. We assume that we are at an $x$ such that <Sign in to see all the formulas> and do the following:

<Sign in to see all the formulas>

Next, let's prove that $G(x)$ is continuous on $[a,b]$. Since $G(x)$ is differentiable on $(a,b)$ we know that it is continuous on $(a,b)$. So now we need to prove that it is continuous from the right at $a$ and continuous from the left at $b$.

We are given that

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and we need to show that

<Sign in to see all the formulas>


First, we know that

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because of this and this.
Therefore, we know that

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Now plugging in

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and

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we get

<Sign in to see all the formulas>

Therefore, $G(x)$ is continuous from the right.

We are given that

<Sign in to see all the formulas>

and we need to show that

<Sign in to see all the formulas>


First, we know that

<Sign in to see all the formulas>

because of this and this.
Therefore, we know that

<Sign in to see all the formulas>

Now plugging in

<Sign in to see all the formulas>

and

<Sign in to see all the formulas>

we get

<Sign in to see all the formulas>

Therefore, $G(x)$ is continuous from the left.

We have now shown that $G(x)$ is continuous on $[a,b]$ and <Sign in to see all the formulas> on $(a,b)$. Therefore, $G(x)$ satisfies the definition of being an antiderivative of $f(x)$ and the proof is complete.

Finally, this proof means that antiderivatives are not unique. In other words, there are many functions that can satisfy the definition of an antiderivative of $f(x)$.