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    Hi. I'm stuck on a particular question and wondered if anyone could help. It is:

    a) Find the first 5 terms in the expansion of (1+2x)^-1


    I can do this, getting the answer 1 - 2x + 4x^2 - 8x^3 + 16x^4

    b) Use the expansion to find an approximation to f 0.1 to -0.2 (1/1+2x) dx.

    c) By evaluating the integral exactly, find an approximation for ln2.


    I can do part a) fine. Also, I can do other approximations using the binomial expansion, just I have no idea where to start when it involves integrals.

    Help would be much appreciated.

    Thanks
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    if the power is 1 the expansion is quite easy....

    the bear
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    (Original post by Marisa_Grace)
    b) Use the expansion to find an approximation to f 0.1 to -0.2 (1/1+2x) dx.
    \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx

    seeing as you've just found the first five terms of the expansion of (1+2x)^{-1}, which is the same as \dfrac{1}{1+2x}, it makes sense to substitute the expansion into the integral. so:

    \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx is approximately equal to \displaystyle\int^{0.1}_{-0.2}1-2x+4x^2-8x^3+16x^4\ dx=...
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    (Original post by the bear)
    if the power is 1 the expansion is quite easy....

    the bear
    Typo lol!
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    (Original post by Pheylan)
    \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx

    seeing as you've just found the first five terms of the expansion of (1+2x)^{-1}, which is the same as \dfrac{1}{1+2x}, it makes sense to substitute the expansion into the integral. so:

    \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx is approximately equal to \displaystyle\int^{0.1}_{-0.2}1-2x+4x^2-8x^3+16x^4\ dx=...
    Thanks. I get that. But how would I do part c)? When I integrate exactly I get, 0.5ln2.
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    (Original post by Marisa_Grace)
    Thanks. I get that. But how would I do part c)? When I integrate exactly I get, 0.5ln2.
    so if \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx=\frac{1}{2}ln2, then \displaystyle ln2=2\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx

    therefore, ln2 is approximately equal to...
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    (Original post by Marisa_Grace)
    When I integrate exactly I get, 0.5ln2.
    so you can equate 0.5ln2 to the value you obtained in part b) ...then double it to find ln2

    the bear

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    (Original post by Pheylan)
    so if \displaystyle\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx=\frac{1}{2}ln2, then \displaystyle ln2=2\int^{0.1}_{-0.2}\frac{1}{1+2x}\ dx

    therefore, ln2 is approximately equal to...
    Thanks again.
 
 
 
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