The second part is relatively simple to prove. When we get the Harmonic Series which . You can construct the following bound to show this:
I feel sorry for your students.Let's light a fire here. No-one knows what school I teach at, right? But if you THINK I might be a teacher at your school, then there is a good chance some questions from this forum might be in your trial. Only 51 pages to sift through.
You're first part is correct, for the second though, I'd advise against the case by case basis, it might work though.
so for we get for some value L.
Now all I need is the case where , but I want to try do a more holistic proof rather than one by cases.
Consider the curve y=1/x^p in the interval [1,infinity) and partition it into rectangles of unit sub-interval, and construct upper bound rectangles.
so for we get for some value L.
Now all I need is the case where , but I want to try do a more holistic proof rather than one by cases.
Constructing rectangle bounds for the curve gives the inequality:
Yep that is pretty much it.Constructing rectangle bounds for the curve gives the inequality:
Consider the RHS, let where
As
There may be an algebraic mistake somewhere as I haven't double checked my working, but it's the same idea nonetheless.
I don't see a need to but feel free to do so if it will help.Can students be expected to use the fundamental theorem of calculus freely?
The first part is taking infinite rectangles between the limits of 1 and 2. Their height will be with width
Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?For something like part i) of your most recent question, you would need to provide the students with a rigorous definition of integration from which to work ...as this is not assumed knowledge in 4U. You also need to state what kind of functions f you want this to be proven for.
(Unless by "prove" you mean "provide a heuristic argument for why this is true using the informal mx2 notion of an integral").
That's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:Is there a problem with the argument; that considering finite n, the sum of the lower rectangles of width 1/n is approximately the area (which is the integral), and then saying as n increases without bound the rectangles all sum up to become the area?
Similar logic is there when deriving integrals for Volumes question, is this not rigorous/not in syllabus ?
Ah alright then :sThat's about as good as you can do with mx2 definitions and notions, but it is certainly not rigorous as the object you are proving things about (the Riemann integral) is not rigorously defined. Other things to think about:
1. What is the definition of area itself? (If you choose to define the integral via area).
2. How do you know that as n->inf the rectangles "area" approaches the "area" under the curve? Beyond the fact that the pictures look more and more alike.
3. You should state an assumption like continuity on f to guarantee those expressions make sense.
The main reason I am stressing these things is that typically, the integral is DEFINED via sums of the sort on the RHS, and "area under curves" is defined via this rigorous notion of integration. This puts things on a solid logical basis.