1. ## MacLAurin Series problem

Using 2 terms of an appropriate MacLaurin series, estimate:

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

I'm not really sure what this is asking me to do. Thanks!

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

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

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

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

4. Showcase22, would you not integrate this function?

$\displaystyle \int_0^1 \frac{1-1+\frac{x^2}{2}}{x}dx=\int_0^1 \frac{x}{2}dx = \frac{1^2}{4}-\frac{0^2}{4} = \frac{1}{4}$

5. Yes, yes you would!

It's early! *sob*
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