What is the integral from 2 to 8 of (e^{1/x})/(x^2) ?

The integral from 2 to 8 of (e^{1/x})/(x^2) is sqrt(e)-\sqrt[8]{e}

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integral from 2 to 8 of (e^{1/x})/(x^2)

\int_{2}^{8} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

e - 8 e

Decimal

0.51557 \dots

Solution steps

\int_{2}^{8} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{1}{8}} - e^{u} d u

= \int_{\frac{1}{2}}^{\frac{1}{8}} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{8}}^{\frac{1}{2}} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{1}{8}}^{\frac{1}{2}} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{\frac{1}{8}}^{\frac{1}{2}})

Simplify

= [e^{u}]_{\frac{1}{8}}^{\frac{1}{2}}

Compute the boundaries:

e - 8 e

= e - 8 e

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What is the integral from 2 to 8 of (e^{1/x})/(x^2) ?
The integral from 2 to 8 of (e^{1/x})/(x^2) is sqrt(e)-\sqrt[8]{e}