Question 1 in Calculus Proof Questions

[Rule]

Since the introductory topic is quite cluttered, it is best to use this topic for attempted solutions to this problem.

Prove that whenever is continuous and , then



(Rule's hints:  Note that this is actually a stronger statement than the Fundamental Theorem of Calculus -- the function f(x) need not be differentiable.  To clarify notation, the function has any positive real number in its domain, and it maps this number to some other real number.  The closed interval a to b is a subset of the open interval from 0 to 1.  Basically a and b are numbers between 0 and 1, and b >= a. So f(x) is not very different from any function you've likely seen so far, except its domain is restricted to the positive reals and doesn't include 0.  Please do not get sidetracked by the notation, as the essence of the proof should be in noting that f is continuous on a bounded interval.  But there are many possible approaches so please keep an open mind and just look at the equation. Also it is not essential that you use the FTC to prove this question, and this result does not trivially follow from the FTC.)


Edit: If you are unfamiliar with some of the notation you may be under the impression that this stuff about f having a domain on an open interval, and so on, is really important to the proof, but it actually isn't.  If you just look at the main equation to be proven, you should be able to start playing with it and get a good idea of how the proof would work. The continuity will probably be important though, at least in some small intuitive way.  Also note that the limit is outside the integral.  In general, limits and integrals cannot be interchanged, although in many introductory courses to calculus this is just done without justification. (If you are interested the function inside the integral must converge uniformly for this operation to be justified, but you do not need to know that for this question).

[Nate]

For any function the integral of the derivative of the function is the function?  What do you want us to prove?  It is obvious.

[Rule]

I had predicted that response, so I thought I gave a fair bit of warning...
See emphasised parts.

(Rule's hints:  Note that this is actually a stronger statement than the Fundamental Theorem of Calculus -- the function f(x) need not be differentiable.  To clarify notation, the function has any positive real number in its domain, and it maps this number to some other real number.  The closed interval a to b is a subset of the open interval from 0 to 1.  Basically a and b are numbers between 0 and 1, and b >= a. So f(x) is not very different from any function you've likely seen so far, except its domain is restricted to the positive reals and doesn't include 0.  Please do not get sidetracked by the notation, as the essence of the proof should be in noting that f is continuous on a bounded interval.  But there are many possible approaches so please keep an open mind and just look at the equation. Also it is not essential that you use the FTC to prove this question, and this result does not trivially follow from the FTC.)


Edit: If you are unfamiliar with some of the notation you may be under the impression that this stuff about f having a domain on an open interval, and so on, is really important to the proof, but it actually isn't.  If you just look at the main equation to be proven, you should be able to start playing with it and get a good idea of how the proof would work. The continuity will probably be important though, at least in some small intuitive way.  Also note that the limit is outside the integral.  In general, limits and integrals cannot be interchanged, although in many introductory courses to calculus this is just done without justification. (If you are interested the function inside the integral must converge uniformly for this operation to be justified, but you do not need to know that for this question).

So, in short, the fundamental theorem of calculus cannot be trivially applied because
1) The function isn't necessarily differentiable
2) Even if f were differentiable, the limit is outside of the integral

I knew of a way of possibly solving this using the FTC, but it is a little work, and is not the simplest solution that I am aware of anyways.  As I said though, the nicest solution to this that I know of is about 1/4 page typed, so it's not too much work.

However, since a similarity between this and the FTC is obvious, it might be worthwhile looking at a good proof of (both parts) of the Fundamental Theorem of Calculus.  It definitely won't give away the solution, but it might give you a few tips.

[Nate]

The integral doesn't depend on h so the limit can be pulled out...

[Rule]

The integral doesn't depend on h so the limit can be pulled out...

The h may be able to come inside, but the limit can't trivially be brought inside.  Here is a counter-example to what you are suggesting:


The left side evaluates to 1/2 while the right side evaluates to 0.  Try working it out for yourself.  The reason both sides are not equivalent is because the sequence of functions being integrated only converges pointwise, not uniformly.