What is the integral from 3 to infinity of (e^{-1/x})/(x^2) ?

The integral from 3 to infinity of (e^{-1/x})/(x^2) is 1-1/(e^{1/3)}

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integral from 3 to infinity of (e^{-1/x})/(x^2)

\int_{3}^{\infty} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

1 - \frac{1}{e ^{\frac{1}{3}}}

Decimal

0.28346 \dots

Solution steps

\int_{3}^{\infty} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Compute the indefinite integral:

\int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x = e^{- \frac{1}{x}} + C

Compute the boundaries:

\int_{3}^{\infty} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x = 1 - \frac{1}{e ^{\frac{1}{3}}}

= 1 - \frac{1}{e ^{\frac{1}{3}}}

What is the integral from 3 to infinity of (e^{-1/x})/(x^2) ?
The integral from 3 to infinity of (e^{-1/x})/(x^2) is 1-1/(e^{1/3)}