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Re: Help for solving this Integrate[Sqrt[t*(1-t)* (t-z),{t,0,z} ] NEW!!!!

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  • Subject: [mg112843] Re: Help for solving this Integrate[Sqrt[t*(1-t)* (t-z),{t,0,z} ] NEW!!!!
  • From: "David Park" <djmpark at comcast.net>
  • Date: Sun, 3 Oct 2010 03:37:58 -0400 (EDT)
  • References: <9668295.21791.1286014260560.JavaMail.root@m06>

I like that a lot and it inspired me to see if we could manipulate the
integral completely by calculation and then evaluate it without ever using
Integrate. Well, anyway this is for those who have Presentations.

<< Presentations` 

We will use the following square root identity at one step.

FullSimplify[
 Sqrt[x y] == Sqrt[-x] Sqrt[-y], {x, y} \[Element] Reals \[And] x < 0]

True

We then use the Student's Integral routines to manipulate the integral to a
simple form.

Print["Initial Integral"]
integrate[Sqrt[t*(1 - z)*(t - z)], {t, 0, z}]
Print["Complete the square on the first and third parts under the \
square root."]
%% // OperateIntegrand[
  MapLevelParts[CompleteTheSquare[#, t] &, {1, {1, 3}}][#] &]
Print["Take out the 1-z factor using the square root identity."]
%% // OperateIntegrand[# /. Sqrt[x_ y_] -> Sqrt[-x] Sqrt[-y] &]
Print["Use a change of variable and simplify."]
%% // ChangeIntegralVariable[u -> 2 t/z - 1, t, 
   DirectSubstitution -> t -> 1/2 (1 + u) z];
% // OperateIntegrand[
  Simplify[#, {v, z} \[Element] Reals \[And] z > 0] &]
Print["Breakout the constant factors."]
step1 = %% // BreakoutIntegral 

This gives us a result that look like:

1/4 Sqrt[-1 + z] z^2 HoldForm[Integrate[Sqrt[1 - u^2], {u, -1, 1}]]

Plotting the integrand in a Manipulate statement we see  that it just sweeps
out the upper half of a unit disk. 

Manipulate[
 Module[{\[Epsilon] = 0.02, ptr, ptu},
  ptr = {u, Sqrt[1 - u^2]};
  ptu = {u, 0};
  Draw2D[
   {{LightOrange, ComplexDisk[0, 1, {0, \[Pi]}]},
    ComplexCircle[0, 1, {0, \[Pi]}],
    Aliasing@Line[{{-1, 0}, {1, 0}}],
    Line[{{0, 0}, ptr, ptu}],
    ComplexText[#1, #1, {0, 1}] & /@ {-1, 0, 1},
    ComplexText["u", u/2, {0, 1}],
    ComplexText["\[DifferentialD]u", u, {0, 1}],
    ComplexText[Sqrt[1 - ("u")^2], u + 1/2 I Sqrt[1 - u^2], {-1.2, 0}],
    Orange,
    Rectangle[{u - \[Epsilon], 0}, {u + \[Epsilon], Sqrt[1 - u^2]}]},
   PlotRange -> {{-1.1, 1.5}, {-0.3, 1.1}},
   PlotRangePadding -> 0.15,
   BaseStyle -> {12},
   ImageSize -> 350]
  ],
 {{u, 0.5}, -1, 1, Appearance -> "Labeled"}] 

We can make the substitution and evaluate the integral without ever using
Integrate. 

step1
% /. integrate[_, _] -> \[Pi]/2 

which gives 1/8 \[Pi] Sqrt[-1 + z] z^2.


David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/  


From: Curtis Osterhoudt [mailto:cfo at lanl.gov] 

In[13]:= Integrate[Sqrt[t*(1 - z)*(t - z)], {t, 0, z}, 
 Assumptions -> z > 0]

Out[13]= 1/8 \[Pi] Sqrt[-1 + z] z^2


It helped me a great deal to plot both the integral and the answers
(especially the integral) vs. t, with various values of z. Then I recognized
the integrand can be cast into a slightly different form:

Integrate[Sqrt[-1*((t - z/2)^2 - z^2/4)], {t, 0, z}, 
 Assumptions -> z > 0 && 0 <= t <= z]

The integral gives 
(\[Pi] z^2)/8

which is simply the area of half a circle with radius z/2; the circles are
obvious if you plot the integrand. They're later scaled by the Sqrt[z-1]
factor. 

On Friday, October 01, 2010 03:41:39 Hugo wrote:
> Could any body help me to solve the following integral in
> mathematica?
> 
> Integrate[sqrt[t * (1-z)*(t-z),{t,0,z}];  t and z are reals; z>0
> 
> I did make a mistake posting the wrong equation yesterday, I apologize
> for that.
> 
> Any help would be appreciate,
> 
> Hugo
> 
> 


-- 
==================================
Curtis Osterhoudt
cfo at remove_this.lanl.and_this.gov
==================================



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