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

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

Popular >

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

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

\frac{e ^{\frac{1}{2}} - 1}{e}

Decimal Notation

0.23865 \dots

Solution steps

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

Apply u-substitution

: \int_{- 1}^{- \frac{1}{2}} e^{u} d u

= \int_{- 1}^{- \frac{1}{2}} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= [e^{u}]_{- 1}^{- \frac{1}{2}}

Compute the boundaries:

\frac{e ^{\frac{1}{2}} - 1}{e}

= \frac{e ^{\frac{1}{2}} - 1}{e}

\int_{0}^{2} \frac{x}{( x + 1 ) ( x + 2 )} d x

\int_{2}^{5} \frac{6 - x}{x ^{3}} d x

\int_{0}^{1} e^{x} cos (x) d x

\int_{10}^{\infty} \frac{1}{x ^{6}} d x

\int_{\frac{π}{2}}^{\frac{5 π}{2}} sin (x) d x

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