Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Find the value of each of the following infinite integrals :
    integrate 1/(4x+1)^2 dx with the limits infinity and 0.
    i have integrated this to get :

    (4x+1)^-1 / -4
    Then,substituting in zero and zero,
    i get :
    -1/4 - - 1/4 = 0.
    but the answer in the back of the book says : 1/4??
    Offline

    7
    ReputationRep:
    (Original post by elvss567)
    Find the value of each of the following infinite integrals :
    integrate 1/(4x+1)^2 dx with the limits infinity and 0.
    i have integrated this to get :

    (4x+1)^-1 / -4
    Then,substituting in zero and zero,
    i get :
    -1/4 - - 1/4 = 0.
    but the answer in the back of the book says : 1/4??

    Integrating gives -1/4*(4x+1)-1 (Agree with you.)

    Therefore the value of the infinite integral is:
    -1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

    Hope this helps.

    Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by JemmaSimmons)
    Integrating gives -1/4*(4x+1)-1 (Agree with you.)

    Therefore the value of the infinite integral is:
    -1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

    Hope this helps.

    Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4
    how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something?
    Offline

    7
    ReputationRep:
    (Original post by elvss567)
    how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something?
    1/2=0.5
    1/1000=0.001
    1/2000000=0.0000005

    Pattern: As you divide 1 by a larger number the result gets smaller. This means that the value of 1/x tends towards 0 as x approaches infinity. So we replace 1/infinity with 0.

    Think of the graph of 1/x.

    Hope this helps.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by JemmaSimmons)
    1/2=0.5
    1/1000=0.001
    1/2000000=0.0000005

    Pattern: As you divide 1 by a larger number the result gets smaller. This means that the value of 1/x tends towards 0 as x approaches infinity. So we say that 1/infinity is 0.

    Think of the graph of 1/x.

    Hope this helps.
    Thankyou!!
    Offline

    7
    ReputationRep:
    (Original post by elvss567)
    Thankyou!!
    No problem.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by JemmaSimmons)
    Integrating gives -1/4*(4x+1)-1 (Agree with you.)

    Therefore the value of the infinite integral is:
    -1/4[1/infinity -1/1)=-1/4[0+-1]= -1/4*-1 = 1/4

    Hope this helps.

    Edit: I took -1/4 out as a factor. Doing it without factorising gives (-1/4*1/infinity) - (-1/4*1)=0--1/4=1/4
    (Original post by elvss567)
    how did you get zero for the infinity bit? what did you substitute in for x? is there an infinity button on the calculator or something?
    Substituting infinity is a very bad practice, infinity isn't even a number you can use in these calculations!

    Essentially you have \displaystyle \int_{0}^{\infty} (4x+1)^{-2} .dx = \lim_{a \rightarrow \infty} \int_{0}^{a} (4x+1)^{-2}

    So  \displaystyle I= \lim_{a \rightarrow \infty} -\frac{1}{4} \left[ \frac{1}{(4x+1)} \right]_0^a

    Thus \displaystyle I= \lim_{a \rightarrow \infty} -\frac{1}{4} \left( \frac{1}{(4a+1)} - \frac{1}{4(0)+1}} \right)

    \displaystyle \Rightarrow I=-\frac{1}{4} \left( \lim_{a \rightarrow \infty} \frac{1}{4a+1} -1 \right)

    Now you need to observe what happens to that term as a goes to infinity, thus what the term's limit is as that happens, then use that value in there.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by JemmaSimmons)
    So we say that 1/infinity is 0.
    No, we say \displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} = 0
    Offline

    7
    ReputationRep:
    (Original post by RDKGames)
    No, we say \displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} = 0
    *Except I have no idea how to type that haha, so just went with the simple, albeit not quite right, version of the explanation. I am also of the mindset that it is best not to sub in infinity as it is not actually a number.

    "This means that the value of 1/x tends towards 0 as x approaches infinity. So we replace 1/infinity with 0. " <-Improved version.
    • Thread Starter
    Offline

    9
    ReputationRep:
    got it! thankyou both!!
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by elvss567)
    got it! thankyou both!!
    Is this definitely core 3? I don't recall limits and improper integrals going into normal maths.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RDKGames)
    Is this definitely core 3? I don't recall limits and improper integrals going into normal maths.
    yep, i took the question straight out of core 3& 4 OCR book by hugh neill and douglas quadling.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 11, 2017

1,652

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
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.