• We have a few events lined up for the October school holidays!
    Watch this space...

Higher Level Integration Marathon & Questions (1 Viewer)

calamebe

Active Member
Joined
Mar 19, 2015
Messages
466
Gender
Male
HSC
2017
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.







Let



where

This is easily evaluated depending on the different cases:

If or , we have , if then and , and if then and .

Case 1:

hence for some constant .

Evaluating the integral for , we obtain that:

when .

Case or .

(which follows from Case 1).

when or .

Case 3:



Let

By symmetry,

Let



By symmetry,

Therefore , so .

Subbing the values back in, we obtain .

Case 4:



This can be done by the same method as case 3, and so .

Hence, when , , and otherwise.
 
Last edited:

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,454
Location
Outside reality
Gender
Male
HSC
2016


Bonus Challenge: Prove the last two integrals are symmetric in (a,b) <=> (b,a) without evaluating it to the final result.
 
Last edited:

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015





______________________________________________



...but unfortunately I don't know how to prove that final integral equals to negative gamma.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
@Paradoxica I think I finally figured what it should've been.



Well, hopefully.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015





Shouldn't be hard assuming I didn't mess up typing the question.
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,454
Location
Outside reality
Gender
Male
HSC
2016





Shouldn't be hard assuming I didn't mess up typing the question.
For all positive real numbers, the iteration derived from Newton's Method applied to the equation q(x)=x²-2 results in the f(x) described above, and converges to the positive real root of the equation.

Hence, fn(x) converges to √ ̅2 for x>0

The point x=0 can be discarded, as it is a set of zero measure.

Hence, the integral is equivalent to



My only problem with this question is that students don't know how to handle problematic points on the domain of an integral.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top