Name:  Jenna Caldwell

Title:    Convergence with ∫(1/x∏)dx for p>1

Description:  This lesson serves to utilize Mathematica's animation
 capabilities to assist students in visualizing the behavior of the classic 
 function 1/x∏ in the first quadrant as p ranges over natural numbers. 
 In particular, it focuses upon how:
       • {x,f(x)} = {1,1}, "  p       
       • f "hugs" the x axis as p increases
       • area under f is finite for p>1 and infinite for p=1, given an upper
        bound Œ ( 0, \ ).
       Too, students will  in general become more familiar with inverse 
       function curves.

Introduction:  Animating the sequence of graphs of 1/x^p as p goes from 1 to a larger integer demonstrates clearly a few of its characteristics as p changes.  Essentially, as p increases, it "pushes" the standard 1/x graph down.

Clear[f,x,a,p]
f[x_,a_]= 1/(x^a);
p[a_]:= Plot[f[x,a], {x,-.0001,3},PlotRange->{{-.001,3},{-.001,3}},
AspectRatio->Automatic,
PlotLabel->"f[x] = (1/x)" ^a];
Show[Table[p[a],{a,1,14,1}]];

Show[%,PlotLabel->"f[x] = (1/x)^p"]

Here the graphs are superimposed onto a single set of axes to demonstrate that, regardless of the exponent p, they all go through a single point.  Why does it do this?

Answer:  Because when x=1, the function f becomes f(1)=1/1^p, and one to the anything is just one, the y value.

Note also how the graphs with larger exponents drop down much faster than 
the familiar 1/x.  It is for this reason that, if we were to measure the area 
under the curve on a interval (a,\), with a>0, we would get a finite value 
for p=2,3,4,... but infinite, or undefined value for p=1.  
I.e., f(x)=1/x∏ DIVERGES for p=1 and CONVERGES for p>1.

In particular, consider how the area under 1/x∏ on the interval (1,100) 
decreases as p increases by evaluating the integral: 
     #åı f(x)dx for (a,b) = (1,100) as p=1,2,3,...13,14

Clear[f,a,b,p]
f[x_] :=1/x
MatrixForm[N[Table[Integrate[f[x]^p, {x, 1,100}],{p,1,14}]]]

These numbers are getting smaller and smaller, which makes sense.