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!
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maybe you can sub the first 2 terms of the following series $\displaystyle cos(x) = 1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$
$\displaystyle \frac{1-1+\frac{x^2}{2}}{x}=\frac{x}{2}$ substituting in x=1 and x=0 gives $\displaystyle \frac{1}{2}-0=\frac{1}{2}$.
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} $
Yes, yes you would! It's early! *sob*
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