Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Hard maths question can you help? watch

    Offline

    14
    ReputationRep:
    (Original post by keromedic)
    Is it 0 or does it just not converge? I thought the former but wolfram outputs the latter.
    Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

    (Original post by Zacken)
    You can't integrate over the discontinuity at x=0, right?
    Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by FireGarden)
    Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

    Isn't the non-thinking answer the one I provided? I.e. 0. As in F(1)-F(-1)=-1+1=0. The graph is symmetrical and because definite integrals are the signed areas, I thought you'd have a net result of 0 because the bits above and below the x-axis would cancel. My bad.

    Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.
    I shall entertain myself by googling some more about discontinuities.
    Offline

    22
    ReputationRep:
    (Original post by keromedic)
    I shall entertain myself by googling some more about discontinuities.
    f(x) = \frac{1}{x^2}, we have \forall x \in \mathbb{R}, x^2 > 0 \Rightarrow \frac{1}{x^2} > 0 \Rightarrow f(x) > 0 for all real x, so the graph always lies above the x-axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

    Furthermore, \int f(x) \, \mathrm{d}x = F(x) = -\frac{1}{x} = -x^{-1}, so F(1) - F(-1) = -\frac{1}{1} - \frac{-1}{-1} = -2.

    I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part.
    Offline

    22
    ReputationRep:
    (Original post by FireGarden)
    Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.

    Well, yes, that's what I was implying by "you can't integrate over the discontinuity" - I'm aware that you can integrate over other, better behaved discontinuities (such as some integrands with trigonometric functions in the denominator), which I learned from hanging around TPIT back in 2013. :rofl:
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by Zacken)
    f(x) = \frac{1}{x^2}, we have \forall x \in \mathbb{R}, x^2 > 0 \Rightarrow \frac{1}{x^2} > 0 \Rightarrow f(x) > 0 for all real x, so the graph always lies above the x-axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

    Furthermore, \int f(x) \, \mathrm{d}x = F(x) = -\frac{1}{x} = -x^{-1}, so F(1) - F(-1) = -\frac{1}{1} - \frac{-1}{-1} = -2.

    I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part.
    I made a mistake
    Offline

    22
    ReputationRep:
    (Original post by keromedic)
    I made a mistake
    Happens to the best of us. :lol:
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by Zacken)
    Happens to the best of us. :lol:
    Forgot that 1*-1=-1 for some reason .

    As for my initial point, it was that I felt that \displaystyle \int_{-1}^1 f(x)dx =\int_{-1}^0 f(x)dx+\int_{0}^1 f(x) dx. It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

    Name:  graph.png
Views: 44
Size:  3.6 KB
    Offline

    22
    ReputationRep:
    (Original post by keromedic)
    Forgot that 1*-1=-1 for some reason .

    As for my initial point, it was that I felt that \displaystyle \int_{-1}^1 f(x)d =\int_{-1}^0 f(x)dx+\int_{0}^1 f(x) dx. It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

    Name:  graph.png
Views: 44
Size:  3.6 KB
    I understand your reasoning, it's just that I'm questioning why your sketch looks like \frac{-1}{x}, \frac{1}{x^2} clearly lies above the x-axis all the way through.
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by Zacken)
    I understand your reasoning, it's just that I'm questioning why your sketch looks like \frac{-1}{x}, \frac{1}{x^2} clearly lies above the x-axis all the way through.
    Because I was wrong. It's been too long.
    Offline

    22
    ReputationRep:
    (Original post by keromedic)
    Because I was wrong. It's been too long.
    Ah, okay. That's fine then. I once had a class test where I wrote x^2 = 4 \Rightarrow x = \pm 4. :facepalm: - silly mistakes, the death of us all. xD
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 2, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.