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    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??
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    (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
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    (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?
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    (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.
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    (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!!
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    (Original post by elvss567)
    Thankyou!!
    No problem.
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    (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.
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    (Original post by JemmaSimmons)
    So we say that 1/infinity is 0.
    No, we say \displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} = 0
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    (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.
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    got it! thankyou both!!
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    (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.
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    (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.
 
 
 
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