integration - Taylor Expansion to Solve Definite Integral within Specific Accuracy - Mathematics Stack Exchange

i have problem asks evaluate 5 decimal places using taylor series definite integral. $$\int_{1}^{2} \frac{e^x}{x} dx$$

i not sure how set up. know taylor expansion

$$\frac{e^x}{x}=\frac{1}{x}+1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^{n-1}}{(n)!}+\frac{x^{n}e^{\xi}}{(n+1)!}$$

so integrate problem , have

$$\left.ln(x)+x+\frac{x^2}{4}+\frac{x^3}{(3!)(3)}+...+\frac{x^{n}}{(n!)(n)}\right|_1^2+\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!} dx$$

now, below taylor remainder of series, , needs upper bounded term have: $$\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!}dx < 5.0x10^{-6}$$.

i have solved 1 of these involved a=0 , b=1 not know how non 0 , 1. means error term is

$$\left.\frac{x^{n+1}e^{\xi}}{(n+1)!(n+1)}\right|_1^2$$

when dealing 0 , 1 error bounded $$\left.\frac{x^{n}e^{\xi}}{(n+1)!(n+1)}\right|_0^1$$ gave

$$\frac{e^{}}{(n+1)!(n+1)}<5.0x10^{-6} $$ meant when n=7 inequality solved took the first 8 terms of taylor series solve within accuracy. however, do when terms below. $$\left.\frac{x^{n+1}e^{\xi}}{(n+1)!(n+1)}\right|_1^2$$

if point me actual inequality trying solve appreciate it. book, since using taylor series $\xi$ has between 0 series starts , x is. problem solved between 0 , 1 , integral between well. however, now, this, between 0 , 2 , bounded 2 larger if making same conjecture. again, @ first thought below inequality solve not sure @ point.

$$\frac{2^{n+1}e^{2}}{(n+1)!(n+1)}-\frac{e}{(n+1)!(n+1)}<5.0x10^{-6}$$ find when n true? n=11

note error term $$\int_{1}^{2}\frac{x^{n}e^{\xi}}{(n+1)!} $$ , want find upper bound it.

the upper bound found multiplying length of interval $2-1 =1$ maximum value of integrand.

your maximum value of integrand $$\frac{2^{n}e^{2}}{(n+1)!} $$ upper upper bound error.

you can continue here.