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Solution
Your input: calculate $$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x}=e^{- \frac{1}{x}}$$$ (for steps, see indefinite integral calculator)
Since there is infinity in the upper bound, this is improper integral of type 1.
To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (infinite).
$$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx=\lim_{x \to \infty}\left(e^{- \frac{1}{x}}\right)-\left(e^{- \frac{1}{x}}\right)|_{\left(x=6\right)}=1 - e^{- \frac{1}{6}}$$$
Answer: $$$\int_{6}^{\infty}\left( \frac{e^{- \frac{1}{x}}}{x^{2}} \right)dx=1 - e^{- \frac{1}{6}}\approx 0.153518275109386$$$