This is a very simple proof.  Suppose that  is an anti-derivative of , i.e. .  Then by the basic properties of derivatives we also have that,

and so  is an anti-derivative of , i.e. .  In other words,

This is also a very simple proof  Suppose that  is an anti-derivative of  and that  is an anti-derivative of .  So we have that  and .  Basic properties of derivatives also tell us that

and so  is an anti-derivative of  and  is an anti-derivative of .    In other words,

From the definition of the definite integral we have,

and we also have,

Therefore,

From the definition of the definite integral we have,

From the definition of the definite integral we have,

Remember that we can pull constants out of summations and out of limits.

First we’ll prove the formula for “+”.  From the definition of the definite integral we have,

To prove the formula for “-” we can either redo the above work with a minus sign instead of a plus sign or we can use the fact that we now know this is true with a plus and using the properties proved above as follows.

If we define  then from the definition of the definite integral we have,

From the definition of the definite integral we have,

Now, by assumption  and we also have  and so we know that

So, from the basic properties of limits we then have,

But the left side is exactly the definition of the integral and so we have,

Since we have  then we know that  on  and so by Property 8  proved above we know that,

We also know from Property 4 that,

So, we then have,

Give  we can use Property 9 on each inequality to write,

Then by Property 7 on the left and right integral to get,

First let’s note that we can say the following about the function and the absolute value,

If we now use Property 9 on each inequality we get,

We know that we can factor the minus sign out of the left integral to get,

Finally, recall that if  then  and of course this works in reverse as well so we then must have,

If  is continuous on [a,b] then,

is continuous on [a,b] and it is differentiable on  and that,

Suppose that x and  are in .  We then have,

Now, using Property 5 of the Integral Properties we can rewrite the first integral and then do a little simplification as follows.

Finally assume that  and we get,

                                                                                       (1)

Let’s now assume that  and since we are still assuming that  are in  we know that   is continuous on  and so by the Extreme Value Theorem we know that there are numbers c and d in  so that  is the absolute minimum of  in  and that  is the absolute maximum of  in .

So, by Property 10 of the Integral Properties we then know that we have,

Or,

Now divide both sides of this by h to get,

 and then use (1) to get,

                                                                                         (2)

Next, if  we can go through the same argument above except we’ll be working on  to arrive at exactly the same inequality above.  In other words, (2) is true provided .

Now, if we take  we also have  and  because both c and d are between x and .  This means that we have the following two limits.

The Squeeze Theorem then tells us that,

                                                                                                (3)

but the left side of this is exactly the definition of the derivative of  and so we get that,

So, we’ve shown that  is differentiable on .

Now, the Theorem at the end of the Definition of the Derivative section tells us that  is also continuous on .   Finally, if we take  or  we can go through a similar argument we used to get  (3) using one-sided limits to get the same result and so the theorem at the end of the Definition of the Derivative section will also tell us that  is continuous at  or  and so in fact  is also continuous on .  

Suppose  is a continuous function on [a,b] and also suppose that  is any anti-derivative for .  Then,

First let  and then we know from Part I of the Fundamental Theorem of Calculus that  and so  is an anti-derivative of  on [a,b].  Further suppose that  is any anti-derivative of  on [a,b] that we want to choose.  So, this means that we must have,

Then, by Fact 2 in the Mean Value Theorem section we know that  and  can differ by no more than an additive constant on .  In other words for  we have,

Now because  and  are continuous on [a,b], if we take the limit of this as  and  we can see that this also holds if  and .

So, for  we know that .  Let’s use this and the definition of  to do the following.

Note that in the last step we used the fact that the variable used in the integral does not matter and so we could change the t’s to x’s.

The average value of a function  over the interval [a,b] is given by,

We know that the average value of n numbers is simply the sum of all the numbers divided by n so let’s start off with that.  Let’s take the interval [a,b] and divide it into n subintervals each of length,

Now from each of these intervals choose the points  and note that it doesn’t really matter how we choose each of these numbers as long as they come from the appropriate interval.  We can then compute the average of the function values  by computing,

                                                                                               (4)

Now, from our definition of  we can get the following formula for n.

and we can plug this into (4) to get,

Let’s now increase n.  Doing this will mean that we’re taking the average of more and more function values in the interval and so the larger we chose n the better this will approximate the average value of the function. 

If we then take the limit as n goes to infinity we should get the average function value.  Or,

We can factor the  out of the limit as we’ve done and now the limit of the sum should look familiar as that is the definition of the definite integral.  So, putting in definite integral we get the formula that we were after.

If  is a continuous function on [a,b] then there is a number c in [a,b] such that,

Let’s start off by defining,

Since  is continuous we know from the Fundamental Theorem of Calculus, Part I that  is continuous on [a,b], differentiable on (a,b) and that .

Now, from the Mean Value Theorem we know that there is a number c such that  and that,

However we know that  and,

So, we then have,

The work done by the force  (assuming that  is continuous) over the range  is,

Let’s start off by dividing the range  into n subintervals of width  and from each of these intervals choose the points .

Now, if n is large and because  is continuous we can assume that  won’t vary by much over each interval and so in the i^th interval we can assume that the force is approximately constant with a value of .  The work on each interval is then approximately,

The total work over  is then approximately,

Finally, if we take the limit of this as n goes to infinity we’ll get the exact work done.  So,

This is, however, nothing more than the definition of the definite integral and so the work done by the force  over  is,