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

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

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

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

e^{2} - e

Decimal

4.67077 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Use the common integral:

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

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

Simplify

= [e^{u}]_{1}^{2}

Compute the boundaries:

e^{2} - e

= e^{2} - e

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