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

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

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

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

\frac{e}{2}

Decimal

0.82436 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{\frac{1}{2}}^{1}

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

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

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

Simplify

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

Compute the boundaries:

\frac{e}{2}

= \frac{e}{2}

\int_{- 2}^{2} (x^{2} - 2) d x

\int_{1}^{2} \frac{x + 3}{x ( x + 2 )} d x

\int_{1}^{5} x^{2} + 2 d x

\int_{0}^{3} (4 t - 1)^{2} d t

\int_{1}^{5} (x^{2} - 1) d x

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