The hard integral thread.

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\displaystyle e\int e^x(1 - \frac{x^2 + 3x + 3}{x^2 + 4x +4}) \ dx

Use \displaystyle to make the expression look a little nicer, as I've done above. :smile:

Have you needed to do any Latex typsetting at Warwick or is that second/third year project stuff?

Reply 41

joostan

Original post by Mihael_Keehl

yep thats what I mean.

I see thanks for the counterexample.

how would you integrate

\displaystyle\int \sec^2(x) \tan^2(x)\ dx

then?

Simple inspection shows that the integral is:

\dfrac{\tan^3(x)}{3}+\mathcal{C}

.
This is essentially an application of the reverse chain rule.

To see this more explicitly you could make a substitution

u=\tan(x)

Reply 42

Mihael_Keehl

Original post by joostan

Simple inspection shows that the integral is:

\dfrac{\tan^3(x)}{3}+\mathcal{C}

.
This is essentially an application of the reverse chain rule.

To see this more explicitly you could make a substitution

u=\tan(x)

I see thank you.

Reply 43

math42

Original post by Zacken

Use \displaystyle to make the expression look a little nicer, as I've done above. :smile:

Have you needed to do any Latex typsetting at Warwick or is that second/third year project stuff?

Thanks for the tip. Nah, we have a second year essay that will require it though.

Reply 44

Zacken

Original post by atsruser

With those limits? Too tired to give this a go at the moment.

Didn't even notice the limits... at least you can still do maths properly when you're tired, unlike me. :facepalm:

Thanks for pointing out. :smile:

Reply 45

atsruser

Original post by Krollo

Spoiler

I'll give you half marks as

1) you don't need a substitution, and
2) you haven't really justified the use of that inequality.

In fact this is an application of the mean value theorem for integrals.

Is there a way to determine this integral exactly, out of interest?

I think not, but Zacken disagrees.

Reply 46

Zacken

Original post by 1 8 13 20 42

Thanks for the tip. Nah, we have a second year essay that will require it though.

Just realised the brackets look funky, use \left( and \right) to automatically fix them.

Ooh, that sounds exciting (and much better than the CATAM stuff at Camb. :tongue:

Reply 47

atsruser

Original post by GuerrillaTactics

Tricky little integral here :colone:

It's certainly hard to read. For us old folk, anyway.

Reply 48

Zacken

Original post by atsruser

I think not, but Zacken disagrees.

Zacken is wrong and really should start reading things properly. :lol:

Reply 49

Louisb19

Original post by Zacken

....

Why was that foreign student guy giving you an attitude in some other thread?

Reply 50

math42

Original post by Zacken

Just realised the brackets look funky, use \left( and \right) to automatically fix them.

Ooh, that sounds exciting (and much better than the CATAM stuff at Camb. :tongue:

I tried but the brackets disappeared. :s-smilie:

Yeah, dunno what I'll do it on though. Maybe Julia sets or something chaos related

Reply 51

Krollo

Original post by atsruser

I'll give you half marks as

1) you don't need a substitution, and
2) you haven't really justified the use of that inequality.

In fact this is an application of the mean value theorem for integrals.

I think not, but Zacken disagrees.

1) I shall have a think...

2) One of those things on my list for things to learn properly lol. Most of my mathematics is held together with blutack and hope rather than rigour...

Posted from TSR Mobile

Reply 52

atsruser

Original post by Zacken

This does, however, bring up the interesting question of: for which (classes of?) functions does the integral equality hold. :-)

Write

y=f(x) \Rightarrow \int (y')^2 dx = y^2 +c \Rightarrow (y')^2 = 2yy'

So for

y' \ne 0

we have

y'=2y \Rightarrow y = ce^{2x}

Reply 53

Zacken

Original post by Louisb19

Why was that foreign student guy giving you an attitude in some other thread?

No clue, man. I've just taken to ignoring his snide comments.

Original post by 1 8 13 20 42

I tried but the brackets disappeared. :s-smilie:

Yeah, dunno what I'll do it on though. Maybe Julia sets or something chaos related

\displaystyle \left( \frac{x^3}{x^2} \right)

seems to work fine for me?

I did an essay on Newtons method for finding complex roots and their basins of attraction, investigated a little into Julia sets. They're fascinating. I enjoyed the topolgy behind it! :smile:

Reply 54

Zacken

Original post by atsruser

Write

y=f(x) \Rightarrow \int (y'[s]wink[/s]^2 dx = y^2 +c \Rightarrow (y'[s]wink[/s]^2 = 2yy'

So for

y' \ne 0

we have

y'=2y \Rightarrow y = ce^{2x}

Niiiiice.

Reply 55

atsruser

Original post by Krollo

Most of my mathematics is held together with blutack and hope rather than rigour...

A bright future in physics awaits ..

Reply 56

atsruser

Original post by 16Characters....

Spoiler

That looks to be right. Very nice.

I was trying to do something similar with an

x \rightarrow -x

sub. but that seems to go nowhere.

Reply 57

16Characters....

Original post by atsruser

That looks to be right. Very nice.

I was trying to do something similar with an

x \rightarrow -x

sub. but that seems to go nowhere.

I am not sure it is creative thought on my part to be honest, I remember a prolonged discussion in the STEP prep thread a few weeks ago about this sort of substitution so it was still at the front of my mind :-)

(edited 8 years ago)

Reply 58

math42

Here's another funny, still pretty transparent one I cooked up which is sort of cheating as the test is not on integration skills (although good knowledge makes that part of it way faster)

\displaystyle \int 2e^x \left(\sum_{n = 0}^{\infty} \frac{x^{4n}}{(4n)!} + \sum_{n = 0}^{\infty} \frac{x^{4n + 1}}{(4n + 1)!} \right)\ dx

Original post by Zacken

No clue, man. I've just taken to ignoring his snide comments.

\displaystyle \left( \frac{x^3}{x^2} \right)

Ah I wasn't including the actual bracket symbol. For what? Or was this a recreational essay? :tongue:

I learnt a lot about them back when preparing for Cambridge interview as something interesting/"supercurricular" to talk about but I've forgotten most of it. :colonhash: