I have to approximate with an error less than 0.1 this integral: $$ \int_1^2 \exp\left(-\frac{1}{x^2}\right)\,dx $$
I understood I have to use the Taylor series, then prove that is uniforme convergence and integrate the "x-dependent" part. Finally I Can use Leibniz to estimate the error (since the series should be a alternate-sign one).
Anyway I am not able to determinate the series since the function is not defined for x_0=0 (so I cannot use McLaurin).
How can I handle this problem? I have a lot of exercises of this kind and I really don't know how to handle them...please help me! Thank you!
Let $z=\frac{1}{x}$: $$\begin{eqnarray*}I=\int_{1}^{2}\exp\left(-\frac{1}{x^2}\right)\, dx = \int_{\frac{1}{2}}^{1}\frac{e^{-z^2}}{z^2}\,dz &=& \sum_{n\geq 0}\frac{(-1)^n}{n!}\int_{1/2}^{1}z^{2n-2}\,dz\\&=&\color{red}{\sum_{n\geq 0}\frac{(-1)^n}{(2n-1)n!}\left(1-\frac{1}{2^{2n-1}}\right)}\end{eqnarray*}$$ The last series converges very fast; moreover, due to alternating signs the approximation error is bounded by the last term considered, and by summing till $n=3$ we get $$ I\approx {\color{red}{\frac{589}{960}}} = 0.6135\ldots $$ with the approximation error being less than $\frac{1}{20}$.