MacLAurin Series problem

**mollymcf2009** · May 26th 2009, 10:16 PM

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!

**pickslides** · May 26th 2009, 10:30 PM

maybe you can sub the first 2 terms of the following series

$cos(x) = 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$

**Showcase_22** · May 26th 2009, 10:41 PM

$\frac{1-1+\frac{x^2}{2}}{x}=\frac{x}{2}$

substituting in x=1 and x=0 gives $\frac{1}{2}-0=\frac{1}{2}$ .

**pickslides** · May 26th 2009, 11:07 PM

Showcase22, would you not integrate this function?

$<br /> \int_0^1 \frac{1-1+\frac{x^2}{2}}{x}dx=\int_0^1 \frac{x}{2}dx = \frac{1^2}{4}-\frac{0^2}{4} = \frac{1}{4}<br />$

**Showcase_22** · May 26th 2009, 11:10 PM

Yes, yes you would!

It's early! *sob*

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