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Math Help - MacLAurin Series problem

  1. #1
    Senior Member mollymcf2009's Avatar
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    MacLAurin Series problem

    Using 2 terms of an appropriate MacLaurin series, estimate:

    \int^{1}_{0} \frac{1-cosx}{x} dx

    I'm not really sure what this is asking me to do. Thanks!
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    maybe you can sub the first 2 terms of the following series

    cos(x) = 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots
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  3. #3
    Super Member Showcase_22's Avatar
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    \frac{1-1+\frac{x^2}{2}}{x}=\frac{x}{2}

    substituting in x=1 and x=0 gives \frac{1}{2}-0=\frac{1}{2}.
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  4. #4
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    Showcase22, would you not integrate this function?

    <br />
\int_0^1 \frac{1-1+\frac{x^2}{2}}{x}dx=\int_0^1 \frac{x}{2}dx = \frac{1^2}{4}-\frac{0^2}{4} = \frac{1}{4}<br />
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  5. #5
    Super Member Showcase_22's Avatar
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    Yes, yes you would!

    It's early! *sob*
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