HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

[Shadowdude]

Re: MX2 Integration Marathon

tis' not integration, no wonder you failed MX2

what

i thought this was the marathon in general


oh oops

 

[deswa1]

Re: MX2 Integration Marathon

what

i thought this was the marathon in general


oh oops

Is that a typo with the du on the end? Regardless, its an impossible integral within elementary functions iirc.

 

[Shadowdude]

Re: MX2 Integration Marathon

Is that a typo with the du on the end? Regardless, its an impossible integral within elementary functions iirc.

the question is correct as typed.

 

[Sindivyn]

Re: MX2 Integration Marathon

what

i thought this was the marathon in general


oh oops

Edit: ucos(x^2) + c?

 

[D94]

Re: MX2 Integration Marathon

Solid question.

Is that a typo with the du on the end? Regardless, its an impossible integral within elementary functions iirc.

Maybe you need to rethink your reasoning as to why it is "impossible". How about:

 

[Sindivyn]

Re: MX2 Integration Marathon

Solid question.



Maybe you need to rethink your reasoning as to why it is "impossible". How about:

I think I see now - the x just becomes a constant (or whatever value is in it)?

 

[D94]

Re: MX2 Integration Marathon

I think I see now - the x just becomes a constant (or whatever value is in it)?

Yep, that's right. A constant with respect to y (in my example).

 

[Shadowdude]

Re: MX2 Integration Marathon

Edit: ucos(x^2) + c?

ding ding ding

correctamundo


now:

 

[Sindivyn]

Re: MX2 Integration Marathon

ding ding ding

correctamundo


now:

let u = x^-1
du = -x^-2dx = -1/x^2 dx

= -

 

[Sindivyn]

Re: MX2 Integration Marathon

Now find

 

[deswa1]

Re: MX2 Integration Marathon

Maybe you need to rethink your reasoning as to why it is "impossible". How about:

Oh yeah I get that. I thought it was a mistype and was meant to be dx- with dx its impossible right?

 

[Shadowdude]

Re: MX2 Integration Marathon

Oh yeah I get that. I thought it was a mistype and was meant to be dx- with dx its impossible right?

No...

You learn how to do it in second year. Uses complex numbers.

Answer is actually

 

[deswa1]

Re: MX2 Integration Marathon

No...

You learn how to do it in second year. Uses complex numbers.

Answer is actually

How do you have a definite value for an indefinite integral?

 

[Shadowdude]

Re: MX2 Integration Marathon

How do you have a definite value for an indefinite integral?

oh oops, limits are 0 to infinity.

LOLOLOLOL

my bad

 

[deswa1]

Re: MX2 Integration Marathon

oh oops, limits are 0 to infinity.

LOLOLOLOL

my bad

Oh right- yeah that makes sense. I've seen some things like this- want to learn how to do them sometime.

 

[Shadowdude]

Re: MX2 Integration Marathon

Oh right- yeah that makes sense. I've seen some things like this- want to learn how to do them sometime.

It's pretty cool how it's derived. Difficult, but cool.

More cool once you look at the graph:

http://www.wolframalpha.com/input/?i=cos%28x^2%29+from+0+to+infinity

It's the area from 0 to whatever the first root is, then minus the area, then plus a little, minus a little... etc. etc. etc. etc.

And it neatly converges.

 

[Aesytic]

Re: MX2 Integration Marathon

Now find

let u=e^x
du = e^x * dx

I = integral{(e^x * e^x)/(e^x+1)}dx
= integral{u/(1+u)}du
= integral{1-(1/(1+u)}du
= u - ln(1+u)
= e^x - ln(e^x+1) + C

 

[nightweaver066]

Re: MX2 Integration Marathon

Oh right- yeah that makes sense. I've seen some things like this- want to learn how to do them sometime.

Try this:

http://community.boredofstudies.org/showthread.php?t=274318&p=6024554&viewfull=1#post6024554

 

[D94]

Re: MX2 Integration Marathon

Oh right- yeah that makes sense. I've seen some things like this- want to learn how to do them sometime.

Probably the Error Function.

 

[Shadowdude]

Re: MX2 Integration Marathon

Well, my method uses integrating over a contour in the complex plane and taking real and imaginary parts.

Which is pretty cool because... well, it is.



if i survive until honours - i might like to dabble in this sort of thing.