$\int_{0}^{1} \frac{\sinh x}{x}\mathrm{d}x$ with an error at most $10^{-1}$.
I tired to find expression of error in order to decide order n, but I can't express it because this Maclaurin series is not alternating series. $$\frac{\sinh x}{x}=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!}$$
How can I estimate this integral within error $10^{-1}$ by using Maclaurin series?
You are using $$ \int_0^1 \frac{\sinh(x)}{x}\; dx = \sum_{n=0}^\infty \int_0^1 \frac{x^{2n}}{(2n+1)!}\; dx = \sum_{n=0}^\infty \frac{1}{(2n+1)(2n+1)!}$$ If you stop at $n=m$, the error will be $$ E_m = \sum_{n=m+1}^\infty \frac{1}{(2n+1)(2n+1)!}$$ Since you only need an error less than $10^{-1}$, you don't have to be very sophisticated in bounding $E$. Take some $m$ for which the first omitted term is less than $1/10$, and see if you can bound the rest by a geometric series.