Prove that if $f(x)$ is continuous on $[a,b]$, then the function \[ F(x)=\int_a^x\,f(u)du \] is continuous on $[a,b]$, differentiable on $(a,b)$ and its derivative is $F\,'(x)=f(x)$ for every $x$ in $(a,b)$.

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Lesson Parent: 
Integration III: Definite Integrals As Upper and Lower Sums
Problem: 

Prove that if $f(x)$ is continuous on $[a,b]$, then the function
\[
F(x)=\int_a^x\,f(u)du
\]
is continuous on $[a,b]$, differentiable on $(a,b)$ and its derivative is $F\,'(x)=f(x)$ for every $x$ in $(a,b)$.

Answer: 

It is true that if $f(x)$ is continuous on $[a,b]$, then the function \[ F(x)=\int_a^x\,f(u)du \] is continuous on $[a,b]$, differentiable on $(a,b)$ and its derivative is $F\,'(x)=f(x)$ for every $x$ in $(a,b)$.

SAMPLE SOLUTION

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Solution: 

Before we start it is important to note that this proof will allow us to say that $F(x)$ is an antiderivative of $f(x)$.

This proof is not overly complicated, but it is long. Here is an outline of the proof:

  1. Pove that <Sign in to see all the formulas> by showing that:
  2. Prove that $F(x)$ is continuous:

Let's first prove that <Sign in to see all the formulas>. Recall from our definition of a derivative that we need to prove that

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To do this we our going to break the proof into proving the left-sided limit and the right-sided limit. We will prove that

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and that

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We know from this lesson that when the left-sided limit and right-sided limits are equal then we have proven the limit.

Since both the left and right-sided limits must exist we will restrict ourselves to $x$ in the open interval $(a,b)$. To understand this more clearly think about what would happen if $x$ were equal to $a$. We know nothing about $f(x)$ to the left of $a$ and therefore cannot determine if

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is true. Similarly, if $x=b$ we could not determine if

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is true because we know nothing about $f(x)$ to the right of $b$.

First, let's prove that

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We know that

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Now because

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we have

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We know the absolute lower and upper bounds on this integral:

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where $m$ and $M$ are the minimum and maximum values of $f(x)$ in the closed interval <Sign in to see all the formulas>. Therefore,

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and by dividing both sides by $h$ we have

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Next, we know that

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and that

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(The use of these two limits is where our assumption that $f(x)$ be continuous are used in this proof.) So from the pinching theorem we know that

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and therefore

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Next, let's prove that

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We will find that the proof for this is nearly identical to the right-sided limit above.

We know that

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Now because

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we have

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Now let's divide both sides by $h$ to find that

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But since $h$ is negative we know that <Sign in to see all the formulas> so we have

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We know the absolute lower and upper bounds on this integral:

<Sign in to see all the formulas>

where $m$ and $M$ are the minimum and maximum values of $f(x)$ in the closed interval <Sign in to see all the formulas>. Now divide both sides by $|h|$ to find that

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But from above we know that

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so we have

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Next, we know that

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and that

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(The use of these two limits is where our assumption that $f(x)$ be continuous are used in this proof.) So from the pinching theorem we know that

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and therefore

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Because we have proven the right and left-sided limits we now know that <Sign in to see all the formulas> in the open interval $(a,b)$.

Now let's prove that $F(x)$ is continuous on $[a,b]$. First, let's recall that because $F(x)$ is differentiable in $(a,b)$ it is also continuous there. So now we need to prove that

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and that

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We know that

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can be put into the equivalent form

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Because

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we can again use upper and lower bounds to find that

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Because of this, this, this, and this we know that

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and

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Finally, the pinching theorem tells is that

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Therefore,

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and $F(x)$ is continuous at $a$ from the right.

We know that

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can be put into the equivalent form

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Because

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we can again use upper and lower bounds to find that

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Because of this, this, this, and this we know that

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and

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Finally, the pinching theorem tells is that

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Therefore,

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and $F(x)$ is continuous at $b$ from the left.

This was a long proof, but we are finally done.