Turn on thread page Beta
    Offline

    12
    ReputationRep:
    (Original post by El Stevo)
    hold on a bit... as x approaches 0 from the negative side, x^-(1/3) starts getting really negative and approaches - 'infinity' as x approaches 0 from the positive side, x^-(1/3) starts getting really huge and approaches +'infinity'

    integrals including singularities can be done using limits, but integrals with asymptotes can't....
    Hmm, INT (1/cbrt(x) ) = 2/3.(x^2/3) + C
    Let F(x) = 2/3(x^2/3)
    INT[-1,1] f(x) = INT[-1,0)f(x) + INT(0,1]f(x)
    = lim(x-> 0-) F(x) - F(-1) + F(1) - lim(x->0+)F(x)
    and lim(x->0) F(x) = 0.
    So I = F(1)-F(-1).
    I'm sure it's right.
    Offline

    1
    ReputationRep:
    umm... Int 1/cbrtx dx = Int x^-(1/3) dx = (3/2)x^(2/3) + c

    i agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.
    Offline

    12
    ReputationRep:
    (Original post by El Stevo)
    umm... Int 1/cbrtx dx = Int x^-(1/3) dx = (3/2)x^(2/3) + c

    i agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.
    , sorry about my mistake
    The area is not only at 1 point or 1 line x = 0. So if e is a very small value, you can assumes area is the integral from -1 to -e, and from e to 1.
    Offline

    1
    ReputationRep:
    i *thought* (my thoughts have been known to be inconsequential blabber before) that that was only the case 'continuous' line with a negligible break in it. as opposed to a line disapearing down and re-emerging from up top?
    Offline

    12
    ReputationRep:
    This similar to integral of the funtion but has only one boundary. Like this integral:
    INT[0, +infinite) of (e^-x). The answer is 1.
    They are all improper integrals.
    Offline

    15
    ReputationRep:
    (Original post by BCHL85)
    Actually, some function can applied it.
    Look at this INT[-1,1] of (1/cbrt(x)). Eventhough 1/cbrt(x) is not continuous function, but it still has INT from -1 to 1.
    The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

    Generally the integral of x^a on (0,1) does exist improperly for a>-1.
    Offline

    1
    ReputationRep:
    (Original post by RichE)
    The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

    Generally the integral of x^a on (0,1) does exist improperly for a>-1.

    Gawd, Riemann, I spit in his general direction! Such a Looooohooooserrrrr! :eek:
    Offline

    12
    ReputationRep:
    (Original post by RichE)
    The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

    Generally the integral of x^a on (0,1) does exist improperly for a>-1.
    Yap, maybe...
    I don't really know about The Fundamental Theorem of Calculus :p:
    Offline

    15
    ReputationRep:
    (Original post by BCHL85)
    Yap, maybe...
    I don't really know about The Fundamental Theorem of Calculus :p:
    The Fundamental Theorem of Calculus just essentially says that integration is the opposite of differentiation; namely it says

    Int[a->b] f'(x) dx = f(b)-f(a)

    But you can't just apply it to any old function like

    f(x) = -1/x and f'(x) = 1/x^2

    - for example you usually insist f'(x) is continuous.

    There are much simpler examples (rather than ones where the function goes off to infinity) where it doesn't work.

    e.g. set

    f(x) = 0 for 0=<x<1 and 1 for 1=<x=<2

    Then f'(x) =0 except at x=1 (but f(x) hasn't zoomed off to infinity, just had a jump) and we see

    0 = Int[0->2] f'(x) dx =/= f(2)-f(0) = 1.
 
 
 
Turn on thread page Beta
Updated: February 10, 2005

University open days

  • Heriot-Watt University
    School of Textiles and Design Undergraduate
    Fri, 16 Nov '18
  • University of Roehampton
    All departments Undergraduate
    Sat, 17 Nov '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 17 Nov '18
Poll
Have you ever experienced bullying?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.