# Approximate with error bounds, the integral $\int^1_0 \dfrac{\sin x}{x}\,dx$

I actually already have the solution to this, but would just like some clarification of how the solution was reached.

The solutions provided used the fact that by Taylor's theorem,

$\sin x = T_6(x) + R_6(x)$

The rest of the solution I understand, but how do you know to use the 6th degree Taylor polynomial? Why not 5th or 7th?

Any explanation would be greatly appreciated! Thanks so much!

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## 2 Answers

If the $6$'th degree polynomial didn't give you a sufficiently small error bound, you might increase the degree until you found one that did.

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The function $\sin[x]$ has a Taylor expansion contains only odd terms, if you have written down the explicit formula you should be able to see everything yourself. The remainder bound should be automatic.

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