Edexcel A2 C4 Mathematics June 2016 - Official Thread

OP

Original post by Zacken

Fam...

\int x^{-2} \, \mathrm{d}x = -x^{-1}

, where'd you get

\ln

from?

Edit to add: Remember that when you integrate, you need to put in modulus terms for logarithms, so even if you managed a logarithmic term, you'd have

\ln |-1| = \ln 1 = 0

(which isn't bad maths at all, there's a justification for us placing moduli on logarithm arguments, btw, if you're interested)

I figured I'd flex my Integration by Parts muscles :cool:

(Positive there's a mistake somewhere as I did it on CodeCogs straight out of my head)... Suffice to say this isn't my brightest hour

u = \frac{1}{x}

,

\frac{dv}{dx} = \frac{1}{x}

\frac{du}{dx}= \frac{-1}{x^2}

,

v = \ln|x|

(edited 8 years ago)

Reply 522

Original post by Euclidean

I figured I'd flex my Integration by Parts muscles :cool:

(Positive there's a mistake somewhere as I did it on CodeCogs straight out of my head)... Suffice to say this isn't my brightest hour

The point of this was that you're meant to get

\displaystyle \int_{-1}^1 x^{-2} \, \mathrm{d}x = -2

and then if you were paying attention, you'd ask yourself why a positive everywhere function was returning a negative area. :tongue:

You can integrate over asymptotes if the function is nicely behaved, by the way, it's not just because of the asymptote that the integral doesn't converge. :-)

Reply 523

OP

Original post by Zacken

The point of this was that you're meant to get

\displaystyle \int_{-1}^1 x^{-2} \, \mathrm{d}x = -2

and then if you were paying attention, you'd ask yourself why a positive everywhere function was returning a negative area. :tongue:

You can integrate over asymptotes if the function is nicely behaved, by the way, it's not just because of the asymptote that the integral doesn't converge. :-)

I failed. But the important thing is, I'm asking the question now :lol:

So you can evaluate some integrals over asymptotes and return with finite values? Seems ludicrous, I'm imagining the Riemann sum over an asymptote and questioning exactly how that works :K:

edit: Is this to do with the fact that the function isn't defined for x=0? Though I can't see how that would result in a negative result..

(edited 8 years ago)

Reply 524

Serine Soul

18

Original post by Zacken

Guess maybe I've been doing too much STEP! :lol:

Cool proof for anybody (else, obviously not you :tongue:

) reading this:

\displaystyle \tan x = \frac{\sin x}{\cos x} = \frac{\cos \left(\frac{\pi}{2} - x\right)}{\sin \left(\frac{\pi}{2} - x\right)} = \cot \left(\frac{\pi}{2} -x\right)

Adding to my notes :yep:

Reply 525

Original post by Euclidean

I failed. But the important thing is, I'm asking the question now :lol:

So you can evaluate some integrals over asymptotes and return with finite values? Seems ludicrous, I'm imagining the Riemann sum over an asymptote and questioning exactly how that works :K:

edit: Is this to do with the fact that the function isn't defined for x=0? Though I can't see how that would result in a negative result..

Try integrating

\displaystyle \int_0^1 \frac{\mathrm{d}x}{\sqrt[3]{x}}

.

Reply 526

Original post by Serine Soul

Adding to my notes :yep:

Short and sweet, my favourite kind! :yep:

Reply 527

OP

Original post by Zacken

Try integrating

\displaystyle \int_0^1 \frac{\mathrm{d}x}{\sqrt[3]{x}}

.

\displaystyle \int_0^1 \frac{\mathrm{d}x}{\sqrt[3]{x}} = \frac{3}{2}

Interesting, the graph tends to

\pm\infty

as x tends to 0 from both directions, so how does that integral of a curve which spans in an infinite direction return a finite value...

And also, why does

\displaystyle \int_{-1}^{1} \frac{1}{x^2} \mathrm{d}x

not

(edited 8 years ago)

Reply 528

Original post by Euclidean

\displaystyle \int_0^1 \frac{\mathrm{d}x}{\sqrt[3]{x}} = \frac{3}{2}

Interesting, the graph tends to

\pm\infty

as x tends to 0 from both directions, so how does that integral of a curve which spans in an infinite direction return a finite value...

Yep, that function is "nicely behaved" enough. :colone:

Reply 529

OP

Original post by Zacken

Yep, that function is "nicely behaved" enough. :colone:

Oh the ambiguity :redface:

Reply 530

Original post by Euclidean

Oh the ambiguity :redface:

Read this for the moment. It's three in the morning right now, but if you remind me some other time, I'll see if I can type up some of my waffling on the matter. :biggrin:

Reply 531

OP

Original post by Zacken

Read this for the moment. It's three in the morning right now, but if you remind me some other time, I'll see if I can type up some of my waffling on the matter. :biggrin:

I gave it a read, I don't understand much of what is going on. The examples clarified the three points though, I'll remind you in the morning as I am very interested to hear more about this :biggrin:

Reply 532

EricPiphany

17

Original post by Euclidean

I failed. But the important thing is, I'm asking the question now :lol:

So you can evaluate some integrals over asymptotes and return with finite values? Seems ludicrous, I'm imagining the Riemann sum over an asymptote and questioning exactly how that works :K:

edit: Is this to do with the fact that the function isn't defined for x=0? Though I can't see how that would result in a negative result..

Riemann sums actually don't converge if the function is unbounded on the interval. In order to define the integral, you have to split the interval up and take limits at every asymptote, like here http://www.sosmath.com/calculus/improper/intro/intro.html
I was thinking a little about this yesterday, the reason the Riemann sum fails seems to be because the Riemann sum allows for overestimation of the area, or more exactly, no matter how small the partition, we can choose a value of the function in the unbounded section as large as we want, making the area as large as we want. But not all this area lies under the curve.

(edited 8 years ago)

Reply 533

Original post by EricPiphany

Riemann sums actually don't converge if the function is unbounded on the interval.

I think that depends on the measure of the set of discontinuities. I recall seeing quite a few Lebesgue integrable functions whose set of discontinuities had measure zero having their Riemann sums converge to certain integrals.

Reply 534

samb1234

13

Original post by Euclidean

Quite possibly one of the best STEP questions ever written :biggrin:

Just a suggestion, I personally don't have an issue with it but you might want to move this conversation. Since this is the C4 thread, some people might not realise that these are questions that are well beyond the level of c4 at times and cause them to worry that they aren't able to answer them. Just something to consider, nice questions though I enjoyed them.

Reply 535

EricPiphany

17

Original post by Zacken

I think that depends on the measure of the set of discontinuities. I recall seeing quite a few Lebesgue integrable functions whose set of discontinuities had measure zero having their Riemann sums converge to certain integrals.

I don't know anything about Lebesgue. I believe unbounded functions aren't Riemann integrable.

Reply 536

edothero

15

Original post by samb1234

Just a suggestion, I personally don't have an issue with it but you might want to move this conversation. Since this is the C4 thread, some people might not realise that these are questions that are well beyond the level of c4 at times and cause them to worry that they aren't able to answer them. Just something to consider, nice questions though I enjoyed them.

All questions here (if not most of them) are accessible with C4 knowledge, by no means are they easy or questions that you'll find in an Edexcel exam, but its good for people to try them. Though I definitely understand where you're coming from

Reply 537