HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Status
Not open for further replies.

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: MX2 2015 Integration Marathon



Start with an IBP
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: MX2 2015 Integration Marathon

Probably was before I had really started being active on this thread.



Bored now. Someone make the questions for me please...
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: MX2 2015 Integration Marathon

For Q4, one way may be to substitute , and note that then . Implementing these substitutions and making some cancellations will give . To evaluate this, we may write . (The purpose of this is to get involved so we can integrate things with tan.) For the , we can use partial fractions. However, this method may be quite tedious and they may be faster methods.
 

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: MX2 2015 Integration Marathon

there is also algebra mistake: not

so I would go

and partial fractions from there
 
Last edited:

porcupinetree

not actually a porcupine
Joined
Dec 12, 2014
Messages
664
Gender
Male
HSC
2015
Re: MX2 2015 Integration Marathon

Can someone give me an overview of how I should approach 'subbing in infinity' like you do while evaluating this integral? e.g. what to do when there's fractions involved, etc
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: MX2 2015 Integration Marathon

Can someone give me an overview of how I should approach 'subbing in infinity' like you do while evaluating this integral? e.g. what to do when there's fractions involved, etc
The rule is to take limits as . The logarithm's argument tends to 1 as , so by continuity of ln, the log tends to ln(1) = 0. Also, the inverse tangent will tend to because . And we use that the limit of a sum is the sum of the limits for continuous functions.

But improper integrals like this wouldn't come up in the HSC probably, and if they did, they'd tell you the rule (i.e. you need to take limits), and they'd probably get you to show what the limits are first.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top